Mathematical induction Questions in English

(A) Let $f(n) = 3^{2n} - 2n + 1$.
For $n = 1$,$f(1) = 3^{2(1)} - 2(1) + 1 = 9 - 2 + 1 = 8$.
For $n = 2$,$f(2) = 3^{2(2)} - 2(2) + 1 = 81 - 4 + 1 = 78$.
Wait,let us re-evaluate the expression.
$3^{2n} = 9^n = (1 + 8)^n = 1 + n(8) + \frac{n(n-1)}{2}(8^2) + \dots$
So,$3^{2n} - 2n + 1 = (1 + 8n + 32n(n-1) + \dots) - 2n + 1 = 6n + 2 + 32n(n-1) + \dots$
Actually,checking $n=1$ gives $8$,which is divisible by $8$.
Checking $n=2$ gives $78$,which is not divisible by $8$.
However,the expression $3^{2n} - 2n + 1$ is always divisible by $8$ if the term was $3^{2n} - 8n + 1$ or similar.
Given the options,let us check $n=1 \implies 8$ and $n=2 \implies 78$.
Since $8$ is divisible by $2$ and $78$ is divisible by $2$,the correct option is $2$.

(B) To determine the correct inequality,we test the first few natural numbers $n \in \{1, 2, 3, \dots\}$.
For $n = 1$: $1 < 2^1$ is $1 < 2$,which is true.
For $n = 2$: $2 < 2^2$ is $2 < 4$,which is true.
For $n = 3$: $3 < 2^3$ is $3 < 8$,which is true.
Since $2^n$ grows exponentially while $n$ grows linearly,$2^n$ will always be greater than $n$ for all natural numbers $n \ge 1$.
Thus,the inequality $n < 2^n$ holds for every natural number $n$.

(C) Let us test the options for $n = 1$:
$(a)$ $2^1 < 1 \implies 2 < 1$ (False)
$(b)$ $1^2 > 2(1) \implies 1 > 2$ (False)
$(c)$ $1^4 < 10^1 \implies 1 < 10$ (True)
$(d)$ $2^{3(1)} > 7(1) + 1 \implies 8 > 8$ (False)
Thus,the correct statement is $n^4 < 10^n$ for $n \in N$.

(B) To determine when ${2^n} < n!$,we test values of $n$ starting from $1$:
For $n = 1$: ${2^1} = 2$,$1! = 1$. Since $2 > 1$,the condition is false.
For $n = 2$: ${2^2} = 4$,$2! = 2$. Since $4 > 2$,the condition is false.
For $n = 3$: ${2^3} = 8$,$3! = 6$. Since $8 > 6$,the condition is false.
For $n = 4$: ${2^4} = 16$,$4! = 24$. Since $16 < 24$,the condition is true.
For $n = 5$: ${2^5} = 32$,$5! = 120$. Since $32 < 120$,the condition is true.
Thus,the inequality ${2^n} < n!$ holds for all integers $n \geq 4$.

(C) To determine the values of $n$ for which ${3^n} > {n^3}$,we test small positive integers:
For $n = 1$: ${3^1} = 3$ and ${1^3} = 1$. Since $3 > 1$,the condition holds.
For $n = 2$: ${3^2} = 9$ and ${2^3} = 8$. Since $9 > 8$,the condition holds.
For $n = 3$: ${3^3} = 27$ and ${3^3} = 27$. Here $27 = 27$,so the condition ${3^n} > {n^3}$ does not hold.
For $n = 4$: ${3^4} = 81$ and ${4^3} = 64$. Since $81 > 64$,the condition holds.
For $n = 5$: ${3^5} = 243$ and ${5^3} = 125$. Since $243 > 125$,the condition holds.
By mathematical induction,it can be shown that for all $n \geq 4$,the inequality ${3^n} > {n^3}$ is true.

(D) The statement $P(n)$ is $n^2 + n$ is odd.
We can write $n^2 + n = n(n + 1)$.
Since $n$ and $n + 1$ are consecutive integers,one of them must be even and the other must be odd.
The product of an even number and an odd number is always even.
Therefore,$n^2 + n$ is always even for all natural numbers $n$.
Since the statement claims $n^2 + n$ is odd,the statement $P(n)$ is false for all $n \in \mathbb{N}$.
Thus,there is no $n$ for which $P(n)$ is true.
Hence,the correct option is $(d)$.

(D) The given statement $P(n) = 2 + 4 + 6 + \dots + 2n$ is an arithmetic series with sum $S_n = n(n + 1)$.
Given the inductive step $P(k) = k(k + 1) + 2 \implies P(k + 1) = (k + 1)(k + 2) + 2$,we are testing the validity of the formula $P(n) = n(n + 1) + 2$.
Since the actual sum of the first $n$ even numbers is $n(n + 1)$,the formula $P(n) = n(n + 1) + 2$ is incorrect for all $n \in N$.
Therefore,the inductive step provided does not prove the formula for any $n$,and thus nothing can be concluded about the validity of the formula $P(n) = n(n + 1) + 2$ from the given information.

(C) We check the statement $P(n)$ for $n = 1$:
$LHS = 1 \times 1! = 1$.
$RHS = (1 + 1)! - 1 = 2! - 1 = 2 - 1 = 1$.
Since $LHS = RHS$,the statement is true for $n = 1$.
Assume the statement is true for $n = k$,i.e.,$1 \times 1! + 2 \times 2! + \dots + k \times k! = (k + 1)! - 1$.
For $n = k + 1$,we have:
$(1 \times 1! + 2 \times 2! + \dots + k \times k!) + (k + 1) \times (k + 1)! = ((k + 1)! - 1) + (k + 1) \times (k + 1)!$.
$= (k + 1)! \times (1 + k + 1) - 1 = (k + 1)! \times (k + 2) - 1 = (k + 2)! - 1$.
This matches the form $(n + 1)! - 1$ for $n = k + 1$.
Thus,by the principle of mathematical induction,the statement is true for all $n \in \mathbb{N}$.

(A) We test the inequality $2^n > 2n + 1$ for natural numbers $n = 1, 2, 3, 4, \dots$
For $n = 1$: $2^1 = 2$ and $2(1) + 1 = 3$. Since $2 > 3$ is false,the inequality does not hold for $n = 1$.
For $n = 2$: $2^2 = 4$ and $2(2) + 1 = 5$. Since $4 > 5$ is false,the inequality does not hold for $n = 2$.
For $n = 3$: $2^3 = 8$ and $2(3) + 1 = 7$. Since $8 > 7$ is true,the inequality holds for $n = 3$.
For $n = 4$: $2^4 = 16$ and $2(4) + 1 = 9$. Since $16 > 9$ is true,the inequality holds for $n = 4$.
By mathematical induction,it can be shown that the inequality holds for all $n \ge 3$.

(D) The principle of mathematical induction states that for a statement $P(n)$ to be true for all natural numbers $n$,two conditions must be satisfied:
$1$. $P(1)$ must be true.
$2$. If $P(k)$ is true,then $P(k+1)$ must be true.
In the given problem,only the implication $P(n) \implies P(n+1)$ is provided. Without the base case (i.e.,$P(1)$ being true),we cannot conclude that $P(n)$ is true for any $n$. Therefore,nothing can be said about the truth of $P(n)$.

(C) Given $S(k) = 1 + 3 + 5 + \dots + (2k - 1) = 3 + k^2$.
For $k = 1$,$S(1) \Rightarrow 1 = 3 + 1^2 = 4$,which is false.
For $k = 2$,$S(2) \Rightarrow 1 + 3 = 3 + 2^2 = 7$,which is false.
Now,assume $S(k)$ is true,i.e.,$1 + 3 + 5 + \dots + (2k - 1) = 3 + k^2$.
Adding $(2(k + 1) - 1) = 2k + 1$ to both sides:
$1 + 3 + 5 + \dots + (2k - 1) + (2k + 1) = 3 + k^2 + 2k + 1 = 3 + (k + 1)^2$.
This shows that $S(k) \Rightarrow S(k + 1)$ is true,even though the base case $S(1)$ is false.

(C) For $n = 1$,we get,
${P^{1 + 1}} + {(P + 1)^{2(1) - 1}} = {P^2} + {(P + 1)^1} = {P^2} + P + 1$.
This expression is clearly divisible by ${P^2} + P + 1$.
Let us assume that the given result is true for $n = m \in N$,i.e.,${P^{m + 1}} + {(P + 1)^{2m - 1}} = k({P^2} + P + 1)$ for some $k \in N$ .....$(i)$
Now,for $n = m + 1$:
${P^{(m + 1) + 1}} + {(P + 1)^{2(m + 1) - 1}} = {P^{m + 2}} + {(P + 1)^{2m + 1}} = {P^{m + 2}} + {(P + 1)^2}{(P + 1)^{2m - 1}}$.
Using equation $(i)$,we substitute ${(P + 1)^{2m - 1}} = k({P^2} + P + 1) - {P^{m + 1}}$:
$= {P^{m + 2}} + {(P + 1)^2}[k({P^2} + P + 1) - {P^{m + 1}}]
= {P^{m + 2}} + {(P + 1)^2} \cdot k({P^2} + P + 1) - {(P + 1)^2}{P^{m + 1}}
= {P^{m + 1}}[P - {(P + 1)^2}] + {(P + 1)^2} \cdot k({P^2} + P + 1)
= {P^{m + 1}}[P - ({P^2} + 2P + 1)] + {(P + 1)^2} \cdot k({P^2} + P + 1)
= {P^{m + 1}}[-{P^2} - P - 1] + {(P + 1)^2} \cdot k({P^2} + P + 1)
= -{P^{m + 1}}({P^2} + P + 1) + {(P + 1)^2} \cdot k({P^2} + P + 1)
= ({P^2} + P + 1)[k{(P + 1)^2} - {P^{m + 1}}]$.
Since this is a multiple of ${P^2} + P + 1$,the result is true for $n = m + 1$.
By the principle of mathematical induction,the result is true for all $n \in N$.

(A) Step $I$: Given the recurrence relation ${U_{n + 1}} = 3{U_n} - 2{U_{n - 1}}$ with ${U_0} = 2$ and ${U_1} = 3$.
For $n = 1$,${U_2} = 3{U_1} - 2{U_0} = 3(3) - 2(2) = 9 - 4 = 5$.
Checking option $(A)$ ${U_n} = {2^n} + 1$:
For $n = 0$,${U_0} = {2^0} + 1 = 1 + 1 = 2$ (True).
For $n = 1$,${U_1} = {2^1} + 1 = 3$ (True).
For $n = 2$,${U_2} = {2^2} + 1 = 5$ (True).
Step $II$: Assume ${U_k} = {2^k} + 1$ and ${U_{k - 1}} = {2^{k - 1}} + 1$.
Step $III$: Using the recurrence relation for $n = k$:
${U_{k + 1}} = 3{U_k} - 2{U_{k - 1}}$
${U_{k + 1}} = 3({2^k} + 1) - 2({2^{k - 1}} + 1)$
${U_{k + 1}} = 3 \cdot {2^k} + 3 - 2 \cdot {2^{k - 1}} - 2$
${U_{k + 1}} = 3 \cdot {2^k} - {2^k} + 1$
${U_{k + 1}} = (3 - 1) \cdot {2^k} + 1 = 2 \cdot {2^k} + 1 = {2^{k + 1}} + 1$.
Thus,by the principle of mathematical induction,${U_n} = {2^n} + 1$ for all $n \in N \cup \{0\}$.

(A) We want to find the smallest $\lambda \in N$ such that $3^n < n!$ for all $n \geq \lambda$.
Check values of $n$:
For $n=1: 3^1 = 3, 1! = 1$. ($3 < 1$ is False)
For $n=2: 3^2 = 9, 2! = 2$. ($9 < 2$ is False)
For $n=3: 3^3 = 27, 3! = 6$. ($27 < 6$ is False)
For $n=4: 3^4 = 81, 4! = 24$. ($81 < 24$ is False)
For $n=5: 3^5 = 243, 5! = 120$. ($243 < 120$ is False)
For $n=6: 3^6 = 729, 6! = 720$. ($729 < 720$ is False)
For $n=7: 3^7 = 2187, 7! = 5040$. ($2187 < 5040$ is True)
Since the inequality holds for $n=7$ and the growth rate of $n!$ is faster than $3^n$,it will hold for all $n \geq 7$.
Thus,the smallest value of $\lambda$ is $7$.

(A) The given statement is $P(n) = n^2 - n + 41$.
For $n = 3$:
$P(3) = 3^2 - 3 + 41 = 9 - 3 + 41 = 47$.
Since $47$ is a prime number,$P(3)$ is true.
For $n = 5$:
$P(5) = 5^2 - 5 + 41 = 25 - 5 + 41 = 61$.
Since $61$ is a prime number,$P(5)$ is true.
Therefore,both $P(3)$ and $P(5)$ are true.

(N/A) Let the given statement be $P(n),$ i.e.,
$P(n): 1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2n+1)}{6}$
For $n=1, P(1): 1^{2} = \frac{1(1+1)(2 \times 1+1)}{6} = \frac{1 \times 2 \times 3}{6} = 1,$ which is true.
Assume that $P(k)$ is true for some positive integer $k,$ i.e.,
$1^{2}+2^{2}+\ldots+k^{2}=\frac{k(k+1)(2k+1)}{6} \quad \dots(1)$
We shall now prove that $P(k+1)$ is also true.
Consider the sum of the first $(k+1)$ terms:
$(1^{2}+2^{2}+\ldots+k^{2})+(k+1)^{2} = \frac{k(k+1)(2k+1)}{6} + (k+1)^{2} \quad [\text{Using } (1)]$
$= \frac{(k+1)[k(2k+1) + 6(k+1)]}{6}$
$= \frac{(k+1)[2k^{2}+k+6k+6]}{6}$
$= \frac{(k+1)[2k^{2}+7k+6]}{6}$
$= \frac{(k+1)(k+2)(2k+3)}{6}$
$= \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \ge 1.$

Let $P(n): 2^n > n$.
Step $1$: For $n = 1$,$2^1 = 2 > 1$. Hence,$P(1)$ is true.
Step $2$: Assume that $P(k)$ is true for some positive integer $k$,i.e.,$2^k > k$ ..............$(1)$.
Step $3$: We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
Multiplying both sides of $(1)$ by $2$,we get:
$2 \cdot 2^k > 2k$
$2^{k+1} > 2k$
Since $k \geq 1$,we have $k + k \geq k + 1$.
Thus,$2k = k + k > k + 1$.
Therefore,$2^{k+1} > k + 1$.
Conclusion: $P(k+1)$ is true whenever $P(k)$ is true. Hence,by the principle of mathematical induction,$P(n)$ is true for every positive integer $n$.

Let $P(n)$ be the statement: $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}$.
Step $1$: For $n = 1$,the $LHS$ is $\frac{1}{1 \times 2} = \frac{1}{2}$ and the $RHS$ is $\frac{1}{1+1} = \frac{1}{2}$. Since $LHS$ = $RHS$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k$,i.e.,$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \ldots + \frac{1}{k(k+1)} = \frac{k}{k+1}$ $(1)$.
Step $3$: We need to prove $P(k+1)$ is true,i.e.,$\frac{1}{1 \times 2} + \ldots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2}$.
Starting from the $LHS$ of $P(k+1)$:
$= \left[ \frac{1}{1 \times 2} + \ldots + \frac{1}{k(k+1)} \right] + \frac{1}{(k+1)(k+2)}$
$= \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$ (Using $(1)$)
$= \frac{k(k+2) + 1}{(k+1)(k+2)} = \frac{k^2 + 2k + 1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$.
Since $P(k+1)$ is true whenever $P(k)$ is true,by the principle of mathematical induction,$P(n)$ is true for all $n \ge 1$.

(N/A) We define the statement $P(n): 7^{n} - 3^{n}$ is divisible by $4.$
Step $1$: Check for $n = 1.$
$P(1): 7^{1} - 3^{1} = 4,$ which is divisible by $4.$ Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k.$
$P(k): 7^{k} - 3^{k} = 4d$ for some integer $d \in \mathbb{N}.$
Step $3$: Prove $P(k+1)$ is true.
$7^{k+1} - 3^{k+1} = 7 \cdot 7^{k} - 3 \cdot 3^{k}$
$= 7 \cdot 7^{k} - 7 \cdot 3^{k} + 7 \cdot 3^{k} - 3 \cdot 3^{k}$
$= 7(7^{k} - 3^{k}) + (7 - 3) \cdot 3^{k}$
$= 7(4d) + 4 \cdot 3^{k}$
$= 4(7d + 3^{k})$
Since $4(7d + 3^{k})$ is a multiple of $4,$ $P(k+1)$ is true.
By the principle of mathematical induction,$7^{n} - 3^{n}$ is divisible by $4$ for every positive integer $n.$

(N/A) Let $P(n)$ be the given statement,i.e.,$P(n): (1 + x)^n \ge (1 + nx)$ for $x > -1.$
We note that $P(n)$ is true when $n = 1,$ since $(1 + x) \ge (1 + x)$ for $x > -1.$
Assume that $P(k): (1 + x)^k \ge (1 + kx)$ for $x > -1$ is true. $(1)$
We want to prove that $P(k + 1)$ is true for $x > -1$ whenever $P(k)$ is true. $(2)$
Consider the identity $(1 + x)^{k+1} = (1 + x)^k(1 + x).$
Given that $x > -1,$ so $(1 + x) > 0.$
Therefore,by using $(1 + x)^k \ge (1 + kx),$ we have $(1 + x)^{k+1} \ge (1 + kx)(1 + x).$
i.e.,$(1 + x)^{k+1} \ge (1 + x + kx + kx^2).$ $(3)$
Here $k$ is a natural number and $x^2 \ge 0,$ so $kx^2 \ge 0.$ Therefore,$(1 + x + kx + kx^2) \ge (1 + x + kx).$
And so we obtain $(1 + x)^{k+1} \ge (1 + (1 + k)x).$
Thus,the statement in $(2)$ is established. Hence,by the principle of mathematical induction,$P(n)$ is true for all natural numbers.

(N/A) Let the statement $P(n)$ be defined as $P(n): 2 \cdot 7^{n} + 3 \cdot 5^{n} - 5$ is divisible by $24$.
Step $1$: For $n=1$,$2 \cdot 7^{1} + 3 \cdot 5^{1} - 5 = 14 + 15 - 5 = 24$,which is divisible by $24$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$2 \cdot 7^{k} + 3 \cdot 5^{k} - 5 = 24q$ for some $q \in N$. This implies $2 \cdot 7^{k} = 24q - 3 \cdot 5^{k} + 5$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$2 \cdot 7^{k+1} + 3 \cdot 5^{k+1} - 5$ is divisible by $24$.
Consider $2 \cdot 7^{k+1} + 3 \cdot 5^{k+1} - 5 = 7(2 \cdot 7^{k}) + 15 \cdot 5^{k} - 5$.
Substitute $2 \cdot 7^{k} = 24q - 3 \cdot 5^{k} + 5$:
$= 7(24q - 3 \cdot 5^{k} + 5) + 15 \cdot 5^{k} - 5$
$= 168q - 21 \cdot 5^{k} + 35 + 15 \cdot 5^{k} - 5$
$= 168q - 6 \cdot 5^{k} + 30$
$= 168q - 6(5^{k} - 5)$.
Since $5^{k} - 5$ is divisible by $4$ for all $k \in N$ (as $5^{k} - 5 = 5(5^{k-1} - 1)$ and $5^{k-1} - 1$ is a multiple of $4$),let $5^{k} - 5 = 4m$.
Then the expression becomes $168q - 6(4m) = 168q - 24m = 24(7q - m)$,which is divisible by $24$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

Let $P(n)$ be the statement: $1^{2} + 2^{2} + \ldots + n^{2} > \frac{n^{3}}{3}$.
Step $1$: For $n = 1$,$1^{2} = 1$ and $\frac{1^{3}}{3} = \frac{1}{3}$. Since $1 > \frac{1}{3}$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$1^{2} + 2^{2} + \ldots + k^{2} > \frac{k^{3}}{3}$ $(1)$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$1^{2} + 2^{2} + \ldots + k^{2} + (k+1)^{2} > \frac{(k+1)^{3}}{3}$.
Starting from the left side:
$1^{2} + 2^{2} + \ldots + k^{2} + (k+1)^{2} > \frac{k^{3}}{3} + (k+1)^{2}$ (using $(1)$)
$= \frac{k^{3} + 3(k^{2} + 2k + 1)}{3} = \frac{k^{3} + 3k^{2} + 6k + 3}{3}$
$= \frac{(k^{3} + 3k^{2} + 3k + 1) + 3k + 2}{3} = \frac{(k+1)^{3} + 3k + 2}{3}$
Since $3k + 2 > 0$ for $k \in N$,we have $\frac{(k+1)^{3} + 3k + 2}{3} > \frac{(k+1)^{3}}{3}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

Let $P(n)$ be the given statement:
$P(n) : (ab)^{n} = a^{n}b^{n}$
Step $1$: For $n = 1$,we have $(ab)^{1} = ab$ and $a^{1}b^{1} = ab$. Since $ab = ab$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k$,i.e.,$(ab)^{k} = a^{k}b^{k}$ .......... $(1)$
Step $3$: We need to prove that $P(k+1)$ is true,i.e.,$(ab)^{k+1} = a^{k+1}b^{k+1}$.
Consider $(ab)^{k+1} = (ab)^{k} \cdot (ab)$.
Using the assumption $(1)$,we get $(ab)^{k+1} = (a^{k}b^{k}) \cdot (ab)$.
By the associative and commutative properties of multiplication,$(ab)^{k+1} = (a^{k} \cdot a) \cdot (b^{k} \cdot b) = a^{k+1}b^{k+1}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true. By the principle of mathematical induction,$P(n)$ is true for all $n \in \mathbb{N}$.

(N/A) Let the given statement be $P(n),$ i.e.,
$P(n): 1+3+3^{2}+\ldots+3^{n-1}=\frac{3^{n}-1}{2}$
For $n=1,$ we have
$P(1): 1 = \frac{3^{1}-1}{2} = \frac{2}{2} = 1,$ which is true.
Assume $P(k)$ is true for some positive integer $k,$ i.e.,
$1+3+3^{2}+\ldots+3^{k-1} = \frac{3^{k}-1}{2}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$1+3+3^{2}+\ldots+3^{k-1}+3^{(k+1)-1}$
$= (1+3+3^{2}+\ldots+3^{k-1}) + 3^{k}$
$= \frac{3^{k}-1}{2} + 3^{k}$ [Using $(i)$]
$= \frac{3^{k}-1 + 2 \cdot 3^{k}}{2}$
$= \frac{(1+2) \cdot 3^{k} - 1}{2}$
$= \frac{3 \cdot 3^{k} - 1}{2}$
$= \frac{3^{k+1}-1}{2}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N.$

Let the given statement be $P(n),$ i.e.,
$P(n): 1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$
For $n=1,$ we have
$P(1): 1^{3}=1=\left(\frac{1(1+1)}{2}\right)^{2}=\left(\frac{2}{2}\right)^{2}=1^{2}=1,$ which is true.
Let $P(k)$ be true for some positive integer $k,$ i.e.,
$1^{3}+2^{3}+3^{3}+\ldots+k^{3}=\left(\frac{k(k+1)}{2}\right)^{2}$ ........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider
$1^{3}+2^{3}+3^{3}+\ldots+k^{3}+(k+1)^{3}$
$= \left(1^{3}+2^{3}+3^{3}+\ldots+k^{3}\right)+(k+1)^{3}$
$= \left(\frac{k(k+1)}{2}\right)^{2}+(k+1)^{3}$ [Using $(i)$]
$= \frac{k^{2}(k+1)^{2}}{4}+(k+1)^{3}$
$= \frac{k^{2}(k+1)^{2}+4(k+1)^{3}}{4}$
$= \frac{(k+1)^{2}\{k^{2}+4(k+1)\}}{4}$
$= \frac{(k+1)^{2}\{k^{2}+4k+4\}}{4}$
$= \frac{(k+1)^{2}(k+2)^{2}}{4}$
$= \left(\frac{(k+1)(k+2)}{2}\right)^{2}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N.$

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+n}=\frac{2n}{n+1}$
For $n=1$,we have
$P(1): 1=\frac{2(1)}{1+1}=\frac{2}{2}=1$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$1+\frac{1}{1+2}+\ldots+\frac{1}{1+2+3+\ldots+k}=\frac{2k}{k+1}$ .........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$S_{k+1} = \left(1+\frac{1}{1+2}+\ldots+\frac{1}{1+2+3+\ldots+k}\right)+\frac{1}{1+2+3+\ldots+k+(k+1)}$
Using $(i)$,we get:
$S_{k+1} = \frac{2k}{k+1} + \frac{1}{\frac{(k+1)(k+2)}{2}}$ [Since $1+2+\ldots+n = \frac{n(n+1)}{2}$]
$S_{k+1} = \frac{2k}{k+1} + \frac{2}{(k+1)(k+2)}$
$S_{k+1} = \frac{2}{(k+1)} \left(k + \frac{1}{k+2}\right)$
$S_{k+1} = \frac{2}{(k+1)} \left(\frac{k^2+2k+1}{k+2}\right)$
$S_{k+1} = \frac{2}{(k+1)} \left(\frac{(k+1)^2}{k+2}\right)$
$S_{k+1} = \frac{2(k+1)}{k+2}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all $n \in N$.

Let the given statement be $P(n)$,i.e.,
$P(n): 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \ldots + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4}$
For $n=1$,we have
$P(1): 1 \cdot 2 \cdot 3 = 6 = \frac{1(1+1)(1+2)(1+3)}{4} = \frac{1 \cdot 2 \cdot 3 \cdot 4}{4} = 6$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \ldots + k(k+1)(k+2) = \frac{k(k+1)(k+2)(k+3)}{4}$ ........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider
$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \ldots + k(k+1)(k+2) + (k+1)(k+2)(k+3)$
$= \{1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \ldots + k(k+1)(k+2)\} + (k+1)(k+2)(k+3)$
$= \frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3)$ [Using $(i)$]
$= (k+1)(k+2)(k+3) \left( \frac{k}{4} + 1 \right)$
$= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$
$= \frac{(k+1)((k+1)+1)((k+1)+2)((k+1)+3)}{4}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.

Let the given statement be $P(n)$,i.e.,
$P(n): 1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \ldots + n \cdot 3^{n} = \frac{(2n - 1) 3^{n+1} + 3}{4}$
For $n = 1$,we have
$P(1): 1 \cdot 3 = 3 = \frac{(2 \cdot 1 - 1) 3^{1+1} + 3}{4} = \frac{3^{2} + 3}{4} = \frac{12}{4} = 3$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \ldots + k \cdot 3^{k} = \frac{(2k - 1) 3^{k+1} + 3}{4}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$(1 \cdot 3 + 2 \cdot 3^{2} + \ldots + k \cdot 3^{k}) + (k+1) \cdot 3^{k+1}$
$= \frac{(2k - 1) 3^{k+1} + 3}{4} + (k+1) 3^{k+1}$ [Using $(i)$]
$= \frac{(2k - 1) 3^{k+1} + 3 + 4(k+1) 3^{k+1}}{4}$
$= \frac{3^{k+1} \{2k - 1 + 4k + 4\} + 3}{4}$
$= \frac{3^{k+1} \{6k + 3\} + 3}{4}$
$= \frac{3^{k+1} \cdot 3 \{2k + 1\} + 3}{4}$
$= \frac{3^{(k+1)+1} \{2(k+1) - 1\} + 3}{4}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all $n \in N$.

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n+1) = \frac{n(n+1)(n+2)}{3}$
For $n=1$,we have:
$P(1): 1 \cdot 2 = 2 = \frac{1(1+1)(1+2)}{3} = \frac{1 \cdot 2 \cdot 3}{3} = 2$,which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + k(k+1) = \frac{k(k+1)(k+2)}{3}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$1 \cdot 2 + 2 \cdot 3 + \ldots + k(k+1) + (k+1)(k+2)$
$= \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$ [Using $(i)$]
$= (k+1)(k+2) \left( \frac{k}{3} + 1 \right)$
$= (k+1)(k+2) \left( \frac{k+3}{3} \right)$
$= \frac{(k+1)(k+2)(k+3)}{3}$
This is the form of $P(k+1)$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}$
For $n = 1$,we have
$P(1): 1 \cdot 3 = 3 = \frac{1(4(1)^2 + 6(1) - 1)}{3} = \frac{4 + 6 - 1}{3} = \frac{9}{3} = 3$,which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots + (2k - 1)(2k + 1) = \frac{k(4k^2 + 6k - 1)}{3}$ $(i)$
We shall now prove that $P(k + 1)$ is true.
Consider the sum up to $(k + 1)$ terms:
$(1 \cdot 3 + 3 \cdot 5 + \ldots + (2k - 1)(2k + 1)) + (2(k + 1) - 1)(2(k + 1) + 1)$
$= \frac{k(4k^2 + 6k - 1)}{3} + (2k + 1)(2k + 3)$ [Using $(i)$]
$= \frac{4k^3 + 6k^2 - k + 3(4k^2 + 8k + 3)}{3}$
$= \frac{4k^3 + 6k^2 - k + 12k^2 + 24k + 9}{3}$
$= \frac{4k^3 + 18k^2 + 23k + 9}{3}$
$= \frac{(k + 1)(4(k + 1)^2 + 6(k + 1) - 1)}{3}$
Thus,$P(k + 1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all $n \in N$.

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): 1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \ldots + n \cdot 2^{n} = (n-1) 2^{n+1} + 2$
For $n=1$,we have
$P(1): 1 \cdot 2 = 2 = (1-1) 2^{1+1} + 2 = 0 + 2 = 2$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \ldots + k \cdot 2^{k} = (k-1) 2^{k+1} + 2$ ........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider
$\{1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \ldots + k \cdot 2^{k}\} + (k+1) \cdot 2^{k+1}$
$= (k-1) 2^{k+1} + 2 + (k+1) 2^{k+1}$
$= 2^{k+1} \{(k-1) + (k+1)\} + 2$
$= 2^{k+1} \cdot (2k) + 2$
$= k \cdot 2^{(k+1)+1} + 2$
$= \{(k+1)-1\} 2^{(k+1)+1} + 2$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.

Let the given statement be $P(n)$,i.e.,
$P(n): \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$
For $n=1$,we have
$P(1): \frac{1}{2}=1-\frac{1}{2^{1}}=\frac{1}{2}$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{k}}=1-\frac{1}{2^{k}}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider
$\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{k}}\right)+\frac{1}{2^{k+1}}$
$= \left(1-\frac{1}{2^{k}}\right)+\frac{1}{2^{k+1}}$ [Using $(i)$]
$= 1 - \frac{1}{2^{k}} + \frac{1}{2^{k+1}}$
$= 1 - \left(\frac{2}{2^{k+1}} - \frac{1}{2^{k+1}}\right)$
$= 1 - \frac{1}{2^{k+1}}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.

Let the given statement be $P(n)$,i.e.,
$P(n): \frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \frac{1}{8 \times 11} + \ldots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4}$
For $n=1$,we have
$P(1): \frac{1}{2 \times 5} = \frac{1}{10} = \frac{1}{6(1)+4} = \frac{1}{10}$,which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$P(k): \frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \ldots + \frac{1}{(3k-1)(3k+2)} = \frac{k}{6k+4}$ ..........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$\frac{1}{2 \times 5} + \ldots + \frac{1}{(3k-1)(3k+2)} + \frac{1}{\{3(k+1)-1\}\{3(k+1)+2\}}$
$= \frac{k}{6k+4} + \frac{1}{(3k+2)(3k+5)}$ [Using $(i)$]
$= \frac{k}{2(3k+2)} + \frac{1}{(3k+2)(3k+5)}$
$= \frac{1}{3k+2} \left( \frac{k}{2} + \frac{1}{3k+5} \right)$
$= \frac{1}{3k+2} \left( \frac{k(3k+5) + 2}{2(3k+5)} \right)$
$= \frac{1}{3k+2} \left( \frac{3k^2 + 5k + 2}{2(3k+5)} \right)$
$= \frac{1}{3k+2} \left( \frac{(3k+2)(k+1)}{2(3k+5)} \right)$
$= \frac{k+1}{2(3k+5)} = \frac{k+1}{6k+10}$
$= \frac{k+1}{6(k+1)+4}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all $n \in N$.

(N/A) Let the given statement be $P(n),$ i.e.,
$P(n): \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \ldots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$
For $n=1,$ we have
$P(1): \frac{1}{1 \cdot 2 \cdot 3} = \frac{1(1+3)}{4(1+1)(1+2)} = \frac{4}{4 \cdot 2 \cdot 3} = \frac{1}{1 \cdot 2 \cdot 3},$ which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \ldots + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)}$ ........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$\left[\frac{1}{1 \cdot 2 \cdot 3} + \ldots + \frac{1}{k(k+1)(k+2)}\right] + \frac{1}{(k+1)(k+2)(k+3)}$
$= \frac{k(k+3)}{4(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)}$ [Using $(i)$]
$= \frac{1}{(k+1)(k+2)} \left\{ \frac{k(k+3)}{4} + \frac{1}{k+3} \right\}$
$= \frac{1}{(k+1)(k+2)} \left\{ \frac{k(k+3)^2 + 4}{4(k+3)} \right\}$
$= \frac{k^3 + 6k^2 + 9k + 4}{4(k+1)(k+2)(k+3)}$
Factoring the numerator $k^3 + 6k^2 + 9k + 4 = (k+1)^2(k+4)$:
$= \frac{(k+1)^2(k+4)}{4(k+1)(k+2)(k+3)} = \frac{(k+1)(k+4)}{4(k+2)(k+3)}$
$= \frac{(k+1)((k+1)+3)}{4((k+1)+1)((k+1)+2)}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N.$

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): a + ar + ar^{2} + \ldots + ar^{n-1} = \frac{a(r^{n} - 1)}{r - 1}$
For $n = 1$,we have
$P(1): a = \frac{a(r^{1} - 1)}{r - 1} = a$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$a + ar + ar^{2} + \ldots + ar^{k-1} = \frac{a(r^{k} - 1)}{r - 1}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum of the first $(k+1)$ terms:
$(a + ar + ar^{2} + \ldots + ar^{k-1}) + ar^{(k+1)-1}$
$= \frac{a(r^{k} - 1)}{r - 1} + ar^{k}$ [Using $(i)$]
$= \frac{a(r^{k} - 1) + ar^{k}(r - 1)}{r - 1}$
$= \frac{ar^{k} - a + ar^{k+1} - ar^{k}}{r - 1}$
$= \frac{ar^{k+1} - a}{r - 1}$
$= \frac{a(r^{k+1} - 1)}{r - 1}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.

Let the given statement be $P(n)$,i.e.,
$P(n): \left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \dots \left(1+\frac{2n+1}{n^{2}}\right)=(n+1)^{2}$
For $n=1$,we have
$P(1): \left(1+\frac{3}{1}\right) = 4 = (1+1)^{2} = 2^{2} = 4$,which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \dots \left(1+\frac{2k+1}{k^{2}}\right)=(k+1)^{2}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the product up to $(k+1)$ terms:
$\left[\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right) \dots \left(1+\frac{2k+1}{k^{2}}\right)\right] \left\{1+\frac{2(k+1)+1}{(k+1)^{2}}\right\}$
$= (k+1)^{2} \left(1+\frac{2k+3}{(k+1)^{2}}\right)$ [Using $(i)$]
$= (k+1)^{2} \left[\frac{(k+1)^{2} + 2k + 3}{(k+1)^{2}}\right]$
$= (k+1)^{2} + 2k + 3$
$= k^{2} + 2k + 1 + 2k + 3 = k^{2} + 4k + 4$
$= (k+2)^{2} = \{(k+1)+1\}^{2}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all $n \in N$.

Let the given statement be $P(n)$,i.e.,
$P(n): \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \dots\left(1+\frac{1}{n}\right)=(n+1)$
For $n=1$,we have
$P(1): \left(1+\frac{1}{1}\right) = 2 = (1+1)$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$P(k): \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \dots\left(1+\frac{1}{k}\right)=(k+1)$ .........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider
$\left[\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \dots\left(1+\frac{1}{k}\right)\right]\left(1+\frac{1}{k+1}\right)$
$= (k+1)\left(1+\frac{1}{k+1}\right)$ [Using $(i)$]
$= (k+1)\left[\frac{(k+1)+1}{k+1}\right]$
$= (k+1)+1$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.

Let the given statement be $P(n)$,i.e.,
$P(n): 1^{2}+3^{2}+5^{2}+\ldots+(2n-1)^{2}=\frac{n(2n-1)(2n+1)}{3}$
For $n=1$,we have
$P(1)=1^{2}=1=\frac{1(2(1)-1)(2(1)+1)}{3}=\frac{1 \times 1 \times 3}{3}=1$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$P(k): 1^{2}+3^{2}+5^{2}+\ldots+(2k-1)^{2}=\frac{k(2k-1)(2k+1)}{3}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$\left\{1^{2}+3^{2}+5^{2}+\ldots+(2k-1)^{2}\right\}+\{2(k+1)-1\}^{2}$
$= \frac{k(2k-1)(2k+1)}{3} + (2k+1)^{2}$ [Using $(i)$]
$= \frac{k(2k-1)(2k+1) + 3(2k+1)^{2}}{3}$
$= \frac{(2k+1)\{k(2k-1) + 3(2k+1)\}}{3}$
$= \frac{(2k+1)\{2k^{2}-k+6k+3\}}{3}$
$= \frac{(2k+1)\{2k^{2}+5k+3\}}{3}$
$= \frac{(2k+1)\{2k^{2}+2k+3k+3\}}{3}$
$= \frac{(2k+1)\{2k(k+1)+3(k+1)\}}{3}$
$= \frac{(2k+1)(k+1)(2k+3)}{3}$
$= \frac{(k+1)\{2(k+1)-1\}\{2(k+1)+1\}}{3}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.

(N/A) Let the given statement be $P(n)$:
$P(n): \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1}$
For $n=1$,we have:
$P(1): \frac{1}{1 \cdot 4} = \frac{1}{4}$ and $\frac{1}{3(1)+1} = \frac{1}{4}$.
Since $\frac{1}{4} = \frac{1}{4}$,$P(1)$ is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$P(k): \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \ldots + \frac{1}{(3k-2)(3k+1)} = \frac{k}{3k+1}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$\left\{ \frac{1}{1 \cdot 4} + \ldots + \frac{1}{(3k-2)(3k+1)} \right\} + \frac{1}{\{3(k+1)-2\}\{3(k+1)+1\}}$
$= \frac{k}{3k+1} + \frac{1}{(3k+1)(3k+4)}$ [Using $(i)$]
$= \frac{1}{3k+1} \left\{ k + \frac{1}{3k+4} \right\}$
$= \frac{1}{3k+1} \left\{ \frac{k(3k+4) + 1}{3k+4} \right\}$
$= \frac{3k^2 + 4k + 1}{(3k+1)(3k+4)}$
$= \frac{(3k+1)(k+1)}{(3k+1)(3k+4)}$
$= \frac{k+1}{3k+4} = \frac{k+1}{3(k+1)+1}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

Let the given statement be $P(n)$:
$P(n): \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \frac{1}{7 \times 9} + \ldots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}$
For $n=1$,we have:
$P(1): \frac{1}{3 \times 5} = \frac{1}{3(2(1)+3)} = \frac{1}{3 \times 5}$,which is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$P(k): \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \ldots + \frac{1}{(2k+1)(2k+3)} = \frac{k}{3(2k+3)}$
We shall now prove that $P(k+1)$ is true.
Consider the sum up to $(k+1)$ terms:
$\left[\frac{1}{3 \times 5} + \ldots + \frac{1}{(2k+1)(2k+3)}\right] + \frac{1}{(2(k+1)+1)(2(k+1)+3)}$
$= \frac{k}{3(2k+3)} + \frac{1}{(2k+3)(2k+5)}$
$= \frac{1}{2k+3} \left[ \frac{k}{3} + \frac{1}{2k+5} \right]$
$= \frac{1}{2k+3} \left[ \frac{k(2k+5) + 3}{3(2k+5)} \right]$
$= \frac{2k^2 + 5k + 3}{3(2k+3)(2k+5)}$
$= \frac{(k+1)(2k+3)}{3(2k+3)(2k+5)}$
$= \frac{k+1}{3(2k+5)} = \frac{k+1}{3(2(k+1)+3)}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all $n \in N$.

Let the given statement be $P(n)$,i.e.,
$P(n): 1+2+3+\ldots+n < \frac{1}{8}(2n+1)^{2}$
Step $1$: Check for $n=1$:
$1 < \frac{1}{8}(2(1)+1)^{2} = \frac{9}{8} = 1.125$
Since $1 < 1.125$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$,i.e.,
$1+2+\ldots+k < \frac{1}{8}(2k+1)^{2}$ ...........$(i)$
Step $3$: Prove $P(k+1)$ is true:
Consider the sum up to $(k+1)$:
$(1+2+\ldots+k) + (k+1) < \frac{1}{8}(2k+1)^{2} + (k+1)$
$= \frac{1}{8} \left\{ (2k+1)^{2} + 8(k+1) \right\}$
$= \frac{1}{8} \left\{ 4k^{2} + 4k + 1 + 8k + 8 \right\}$
$= \frac{1}{8} \left\{ 4k^{2} + 12k + 9 \right\}$
$= \frac{1}{8} (2k+3)^{2}$
$= \frac{1}{8} \{2(k+1)+1\}^{2}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): n(n+1)(n+5)$ is a multiple of $3$.
Step $1$: For $n=1$,$1(1+1)(1+5) = 1(2)(6) = 12$,which is a multiple of $3$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$,i.e.,$k(k+1)(k+5) = 3m$ for some $m \in N$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$(k+1)(k+2)(k+6)$ is a multiple of $3$.
Consider $(k+1)(k+2)(k+6) = (k+1)(k+2)(k+5+1)$
$= (k+1)(k+2)(k+5) + (k+1)(k+2)$
$= (k+1)(k+5)(k+2) + (k^2+3k+2)$
Since $k(k+1)(k+5) = 3m$,we have $(k+1)(k+5) = \frac{3m}{k}$. This approach is complex,so let's expand differently:
$(k+1)(k+2)(k+6) = (k^2+3k+2)(k+6) = k^3 + 6k^2 + 3k^2 + 18k + 2k + 12$
$= k^3 + 9k^2 + 20k + 12$
$= (k^3 + 6k^2 + 5k) + (3k^2 + 15k + 12)$
$= k(k^2 + 6k + 5) + 3(k^2 + 5k + 4)$
$= k(k+1)(k+5) + 3(k^2 + 5k + 4)$
Since $k(k+1)(k+5)$ is a multiple of $3$ (by assumption) and $3(k^2+5k+4)$ is clearly a multiple of $3$,their sum is also a multiple of $3$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): 10^{2n-1} + 1$ is divisible by $11$.
Step $1$: Check for $n = 1$.
$P(1) = 10^{2(1)-1} + 1 = 10^1 + 1 = 11$,which is divisible by $11$.
Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$.
$10^{2k-1} + 1 = 11m$,where $m \in N$ --- $(i)$
Step $3$: Prove $P(k+1)$ is true.
Consider $10^{2(k+1)-1} + 1 = 10^{2k+2-1} + 1 = 10^{2k+1} + 1$.
$= 10^2 \cdot 10^{2k-1} + 1$
$= 100 \cdot (11m - 1) + 1$ [From $(i)$,$10^{2k-1} = 11m - 1$]
$= 1100m - 100 + 1$
$= 1100m - 99$
$= 11(100m - 9)$.
Since $11(100m - 9)$ is a multiple of $11$,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

(N/A) Let the given statement be $P(n)$,i.e.,$P(n): x^{2n}-y^{2n}$ is divisible by $x+y$.
Step $1$: Check for $n=1$.
$P(1): x^{2(1)}-y^{2(1)} = x^2-y^2 = (x+y)(x-y)$,which is clearly divisible by $(x+y)$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$,i.e.,$x^{2k}-y^{2k} = m(x+y)$ for some integer $m$. .......$(i)$
Step $3$: Prove $P(k+1)$ is true.
Consider $x^{2(k+1)}-y^{2(k+1)} = x^{2k} \cdot x^2 - y^{2k} \cdot y^2$.
$= x^2(x^{2k} - y^{2k} + y^{2k}) - y^{2k} \cdot y^2$
$= x^2(x^{2k} - y^{2k}) + y^{2k}(x^2 - y^2)$
$= x^2[m(x+y)] + y^{2k}(x+y)(x-y)$ [Using $(i)$]
$= (x+y)[m x^2 + y^{2k}(x-y)]$.
Since $(x+y)$ is a factor,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

(N/A) Let the given statement be $P(n)$,i.e.,$P(n): 3^{2n+2} - 8n - 9$ is divisible by $8$.
Step $1$: For $n = 1$,$3^{2(1)+2} - 8(1) - 9 = 3^4 - 8 - 9 = 81 - 17 = 64$. Since $64$ is divisible by $8$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$,i.e.,$3^{2k+2} - 8k - 9 = 8m$ for some $m \in N$. Thus,$3^{2k+2} = 8m + 8k + 9$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$3^{2(k+1)+2} - 8(k+1) - 9$ is divisible by $8$.
Consider $3^{2k+4} - 8k - 8 - 9 = 3^2 \cdot 3^{2k+2} - 8k - 17$.
Substituting $3^{2k+2} = 8m + 8k + 9$:
$= 9(8m + 8k + 9) - 8k - 17$
$= 72m + 72k + 81 - 8k - 17$
$= 72m + 64k + 64$
$= 8(9m + 8k + 8)$.
Since this is a multiple of $8$,$P(k+1)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): 41^{n}-14^{n}$ is a multiple of $27$.
For $n=1$:
$41^{1}-14^{1} = 27$,which is a multiple of $27$.
Thus,$P(1)$ is true.
Assume $P(k)$ is true for some positive integer $k$,i.e.,
$41^{k}-14^{k} = 27m$ for some $m \in N$ ........$(i)$
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
Consider $41^{k+1}-14^{k+1}$:
$= 41 \cdot 41^{k} - 14 \cdot 14^{k}$
$= 41(41^{k} - 14^{k} + 14^{k}) - 14 \cdot 14^{k}$
$= 41(27m) + 41 \cdot 14^{k} - 14 \cdot 14^{k}$
$= 41 \cdot 27m + 14^{k}(41 - 14)$
$= 41 \cdot 27m + 27 \cdot 14^{k}$
$= 27(41m + 14^{k})$
$= 27r$,where $r = (41m + 14^{k})$ is a natural number.
Therefore,$41^{k+1}-14^{k+1}$ is a multiple of $27$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): (2n + 7) < (n + 3)^{2}$
Step $1$: Check for $n = 1$.
$2(1) + 7 = 9$ and $(1 + 3)^{2} = 16$.
Since $9 < 16$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$,i.e.,
$(2k + 7) < (k + 3)^{2}$ ..........$(i)$
Step $3$: We shall now prove that $P(k + 1)$ is true whenever $P(k)$ is true.
Consider the expression for $n = k + 1$:
$2(k + 1) + 7 = (2k + 7) + 2$
Using $(i)$,we have:
$2(k + 1) + 7 < (k + 3)^{2} + 2$
$2(k + 1) + 7 < k^{2} + 6k + 9 + 2$
$2(k + 1) + 7 < k^{2} + 6k + 11$
Since $k^{2} + 6k + 11 < k^{2} + 8k + 16$ for all $k \in N$ (because $2k + 5 > 0$ for $k \geq 1$),
$2(k + 1) + 7 < (k + 4)^{2}$
$2(k + 1) + 7 < \{(k + 1) + 3\}^{2}$
Thus,$P(k + 1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.

(N/A) Let $P(n)$ be the statement: $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \ldots + \frac{1}{(2n-1)(2n+1)} = \frac{n}{2n+1}$.
Step $1$: For $n = 1$,$LHS$ $= \frac{1}{1 \cdot 3} = \frac{1}{3}$ and $RHS$ $= \frac{1}{2(1)+1} = \frac{1}{3}$. Since $LHS$ $=$ $RHS$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$\frac{1}{1 \cdot 3} + \ldots + \frac{1}{(2k-1)(2k+1)} = \frac{k}{2k+1}$.
Step $3$: For $n = k+1$,we need to show $P(k+1)$ is true,i.e.,$\frac{1}{1 \cdot 3} + \ldots + \frac{1}{(2k+1)(2k+3)} = \frac{k+1}{2k+3}$.
$LHS$ $= \left(\frac{1}{1 \cdot 3} + \ldots + \frac{1}{(2k-1)(2k+1)}\right) + \frac{1}{(2k+1)(2k+3)}$
$= \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)}$
$= \frac{k(2k+3) + 1}{(2k+1)(2k+3)} = \frac{2k^2 + 3k + 1}{(2k+1)(2k+3)}$
$= \frac{(2k+1)(k+1)}{(2k+1)(2k+3)} = \frac{k+1}{2k+3} = \text{RHS}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true. By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

(N/A) Let $P(n): 1+2+2^2+\ldots+2^n = 2^{n+1}-1$ for all $n \in N$.
Step $I$: For $n=1$,
$LHS$ $= 1+2^1 = 3$.
$RHS$ $= 2^{1+1}-1 = 2^2-1 = 4-1 = 3$.
Since $LHS$ $= RHS$,the statement is true for $n=1$.
Step $II$: Assume $P(k)$ is true for some $k \in N$,i.e.,$1+2+2^2+\ldots+2^k = 2^{k+1}-1$.
Step $III$: For $n=k+1$,we need to show $P(k+1)$ is true,i.e.,$1+2+2^2+\ldots+2^k+2^{k+1} = 2^{(k+1)+1}-1 = 2^{k+2}-1$.
$LHS$ $= (1+2+2^2+\ldots+2^k) + 2^{k+1}$.
Using the assumption from Step $II$,$LHS$ $= (2^{k+1}-1) + 2^{k+1}$.
$= 2 \times 2^{k+1} - 1 = 2^{k+2}-1$.
Since $LHS$ $= RHS$,the statement is true for $n=k+1$.
By the principle of mathematical induction,the statement is true for all $n \in N$.

Let $P(n)$ be the statement: $\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right) \ldots \left(1-\frac{1}{n^{2}}\right)=\frac{n+1}{2n}$.
Step $1$: For $n=2$,the $LHS$ is $\left(1-\frac{1}{2^{2}}\right) = 1-\frac{1}{4} = \frac{3}{4}$.
The $RHS$ is $\frac{2+1}{2(2)} = \frac{3}{4}$.
Since $LHS$ = $RHS$,$P(2)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \geq 2$,i.e.,$\left(1-\frac{1}{2^{2}}\right) \ldots \left(1-\frac{1}{k^{2}}\right) = \frac{k+1}{2k}$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$\left(1-\frac{1}{2^{2}}\right) \ldots \left(1-\frac{1}{k^{2}}\right)\left(1-\frac{1}{(k+1)^{2}}\right) = \frac{(k+1)+1}{2(k+1)} = \frac{k+2}{2(k+1)}$.
Using the assumption from Step $2$:
$LHS$ = $\left(\frac{k+1}{2k}\right) \left(1-\frac{1}{(k+1)^{2}}\right) = \left(\frac{k+1}{2k}\right) \left(\frac{(k+1)^{2}-1}{(k+1)^{2}}\right)$.
$= \left(\frac{k+1}{2k}\right) \left(\frac{k^{2}+2k+1-1}{(k+1)^{2}}\right) = \left(\frac{k+1}{2k}\right) \left(\frac{k(k+2)}{(k+1)^{2}}\right)$.
$= \frac{k+2}{2(k+1)}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true. By the principle of mathematical induction,$P(n)$ is true for all $n \geq 2$.