Mathematical induction Questions in English

(N/A) Let $P(n)$ be the statement: $3 \times 6 + 6 \times 9 + 9 \times 12 + \ldots + (3n)(3n + 3) = 3n(n + 1)(n + 2)$.
Step $1$: For $n = 1$,the $LHS$ is $3 \times 6 = 18$. The $RHS$ is $3(1)(1 + 1)(1 + 2) = 3 \times 2 \times 3 = 18$. Since $LHS$ = $RHS$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$3 \times 6 + 6 \times 9 + \ldots + (3k)(3k + 3) = 3k(k + 1)(k + 2)$.
Step $3$: We need to show $P(k + 1)$ is true,i.e.,$3 \times 6 + \ldots + (3k)(3k + 3) + (3(k + 1))(3(k + 1) + 3) = 3(k + 1)(k + 2)(k + 3)$.
Adding $(3(k + 1))(3(k + 1) + 3)$ to both sides of the assumption:
$LHS$ = $3k(k + 1)(k + 2) + (3k + 3)(3k + 6)$
= $3k(k + 1)(k + 2) + 3(k + 1) \times 3(k + 2)$
= $3(k + 1)(k + 2) [k + 3]$
= $3(k + 1)(k + 2)(k + 3)$.
Thus,$P(k + 1)$ is true. By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

(A) Let $P(n)$ be the statement: $a+(a+d)+(a+2d)+\ldots+(a+(n-1)d) = \frac{n}{2}[2a+(n-1)d]$.
Step $1$: For $n=1$,the $LHS$ is $a$ and the $RHS$ is $\frac{1}{2}[2a+(1-1)d] = \frac{1}{2}(2a) = a$. Since $LHS$ = $RHS$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$a+(a+d)+\ldots+(a+(k-1)d) = \frac{k}{2}[2a+(k-1)d]$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$a+(a+d)+\ldots+(a+(k-1)d) + (a+kd) = \frac{k+1}{2}[2a+kd]$.
Starting from the $LHS$ of $P(k+1)$:
$= \frac{k}{2}[2a+(k-1)d] + (a+kd)$
$= \frac{2ak + k(k-1)d + 2a + 2kd}{2}$
$= \frac{2a(k+1) + (k^2-k+2k)d}{2}$
$= \frac{2a(k+1) + (k^2+k)d}{2}$
$= \frac{2a(k+1) + k(k+1)d}{2}$
$= \frac{k+1}{2}[2a+kd]$.
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 $4+8+12+\ldots+4n = 2n(n+1)$.
Step $1$: For $n=1$,the left-hand side is $4(1) = 4$ and the right-hand side is $2(1)(1+1) = 2(2) = 4$.
Since $4=4$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$4+8+12+\ldots+4k = 2k(k+1)$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$4+8+12+\ldots+4k+4(k+1) = 2(k+1)(k+2)$.
Starting with the left-hand side:
$(4+8+12+\ldots+4k) + 4(k+1) = 2k(k+1) + 4(k+1)$
$= 2k^2 + 2k + 4k + 4$
$= 2k^2 + 6k + 4$
$= 2(k^2 + 3k + 2)$
$= 2(k+1)(k+2)$.
This matches the right-hand side for $n=k+1$.
Thus,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

(N/A) Let $P(n): 7^{n}-3^{n}$ is divisible by $4$.
Step $1$: For $n=1$,$7^{1}-3^{1} = 4$,which is divisible by $4$. So,$P(1)$ is true.
Step $2$: Assume $P(m)$ is true for some $m \in N$,i.e.,$7^{m}-3^{m} = 4k$ for some integer $k$. Thus,$7^{m} = 4k + 3^{m}$.
Step $3$: For $n=m+1$,we have $7^{m+1}-3^{m+1}$.
$= 7 \cdot 7^{m} - 3 \cdot 3^{m}$
$= 7(4k + 3^{m}) - 3 \cdot 3^{m}$
$= 28k + 7 \cdot 3^{m} - 3 \cdot 3^{m}$
$= 28k + 4 \cdot 3^{m}$
$= 4(7k + 3^{m})$.
Since $4(7k + 3^{m})$ is a multiple of $4$,$P(m+1)$ is true.
By the principle of mathematical induction,$7^{n}-3^{n}$ is divisible by $4$ for all $n \in N$.

(N/A) Let $P(n): 2^{3n}-1$ be divisible by $7$.
Step $1$: Base case for $n=1$:
$P(1) = 2^{3(1)} - 1 = 8 - 1 = 7$,which is divisible by $7$.
So,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$:
$P(k) = 2^{3k} - 1 = 7m$ for some integer $m$.
Therefore,$2^{3k} = 7m + 1$ --- $(1)$
Step $3$: Prove $P(k+1)$ is true:
$P(k+1) = 2^{3(k+1)} - 1 = 2^{3k+3} - 1$
$= 2^{3k} \times 2^3 - 1$
$= (7m + 1) \times 8 - 1$ (using equation $(1)$)
$= 56m + 8 - 1$
$= 56m + 7$
$= 7(8m + 1)$
Since $7(8m + 1)$ is a multiple of $7$,$P(k+1)$ is divisible by $7$.
Thus,by the principle of mathematical induction,$2^{3n}-1$ is divisible by $7$ for all $n \in N$.

Let $P(n)$ be the statement that $3^{2n} - 1$ is divisible by $8$.
Step $1$: For $n = 1$,$3^{2(1)} - 1 = 9 - 1 = 8$,which is divisible by $8$. Thus,$P(1)$ is true.
Step $2$: Assume $P(m)$ is true for some $m \in N$,i.e.,$3^{2m} - 1 = 8k$ for some integer $k$. So,$3^{2m} = 8k + 1$.
Step $3$: We need to show $P(m+1)$ is true,i.e.,$3^{2(m+1)} - 1$ is divisible by $8$.
$3^{2(m+1)} - 1 = 3^{2m} \times 3^2 - 1$
$= (8k + 1) \times 9 - 1$
$= 72k + 9 - 1$
$= 72k + 8$
$= 8(9k + 1)$.
Since $8(9k + 1)$ is divisible by $8$,$P(m+1)$ is true.
By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

Let $P(n): 4^{n}-1$ is divisible by $3$ for all $n \in \mathbb{N}$.
Step $1$: For $n=1$,$P(1) = 4^{1}-1 = 3$,which is divisible by $3$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in \mathbb{N}$,i.e.,$4^{k}-1 = 3m$ for some integer $m \in \mathbb{N}$. This implies $4^{k} = 3m+1$ $(i)$.
Step $3$: We need to prove $P(k+1)$ is true,i.e.,$4^{k+1}-1$ is divisible by $3$.
Consider $4^{k+1}-1 = 4 \cdot 4^{k}-1$.
Substituting from $(i)$,we get $4(3m+1)-1 = 12m+4-1 = 12m+3 = 3(4m+1)$.
Since $3(4m+1)$ is a multiple of $3$,$4^{k+1}-1$ is divisible by $3$.
Conclusion: By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.

(N/A) Let $P(n): 2^{3n} - 1$ be divisible by $7$ for all $n \in \mathbb{N}$.
Step $1$: For $n = 1$,$P(1): 2^{3(1)} - 1 = 8 - 1 = 7$,which is divisible by $7$.
Therefore,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in \mathbb{N}$.
That is,$2^{3k} - 1 = 7m$ for some $m \in \mathbb{N}$,which implies $2^{3k} = 7m + 1$.
Step $3$: We need to show $P(k+1)$ is true.
$P(k+1): 2^{3(k+1)} - 1 = 2^{3k} \cdot 2^3 - 1$.
Substituting $2^{3k} = 7m + 1$:
$= (7m + 1) \cdot 8 - 1$
$= 56m + 8 - 1$
$= 56m + 7$
$= 7(8m + 1)$,which is clearly divisible by $7$.
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 $P(n): n^{3}-7n+3$ be divisible by $3$ for all $n \in N$.
Step $1$: For $n=1$,$P(1) = (1)^{3}-7(1)+3 = 1-7+3 = -3$.
Since $-3$ is divisible by $3$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$k^{3}-7k+3 = 3m$ for some integer $m$.
Step $3$: We need to show $P(k+1)$ is true.
$P(k+1) = (k+1)^{3}-7(k+1)+3$
$= (k^{3}+3k^{2}+3k+1) - 7k - 7 + 3$
$= (k^{3}-7k+3) + 3k^{2}+3k-6$
$= 3m + 3(k^{2}+k-2)$
$= 3(m+k^{2}+k-2)$.
Since $3(m+k^{2}+k-2)$ is divisible by $3$,$P(k+1)$ is true.
Thus,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.

(N/A) Let $P(n): 3^{2n} - 1$ is divisible by $8$.
Step $1$: For $n = 1$,$P(1) = 3^{2(1)} - 1 = 9 - 1 = 8$,which is divisible by $8$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k$,i.e.,$3^{2k} - 1 = 8m$ for some integer $m$. This implies $3^{2k} = 8m + 1$ $(i)$.
Step $3$: For $n = k + 1$,we need to show $P(k + 1)$ is true,i.e.,$3^{2(k+1)} - 1$ is divisible by $8$.
$3^{2(k+1)} - 1 = 3^{2k+2} - 1 = 3^{2k} \cdot 3^2 - 1 = 9 \cdot 3^{2k} - 1$.
Substituting $(i)$,we get $9(8m + 1) - 1 = 72m + 9 - 1 = 72m + 8 = 8(9m + 1)$.
Since $8(9m + 1)$ is divisible by $8$,$P(k + 1)$ is true.
Hence,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.

(N/A) Let $P(n): 7^{n}-2^{n}$ be divisible by $5$.
For $n=1$:
$P(1): 7^{1}-2^{1} = 7-2 = 5$,which is divisible by $5$.
Therefore,$P(1)$ is true.
Assume that $P(k)$ is true for some $k \in N$:
$P(k): 7^{k}-2^{k} = 5m$,where $m \in N$ (Equation $i$).
We need to prove that $P(k+1)$ is true:
$P(k+1): 7^{k+1}-2^{k+1} = 7 \cdot 7^{k} - 2 \cdot 2^{k}$
$= 7 \cdot 7^{k} - 7 \cdot 2^{k} + 7 \cdot 2^{k} - 2 \cdot 2^{k}$
$= 7(7^{k}-2^{k}) + 2^{k}(7-2)$
$= 7(5m) + 2^{k}(5)$
$= 5(7m + 2^{k})$
Since $5(7m + 2^{k})$ is a multiple of $5$,$P(k+1)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.

(N/A) Let $P(n): x^{n}-y^{n}$ is divisible by $(x-y)$ for all $n \in \mathbb{N}$.
Step $1$: For $n=1$,$P(1): x^{1}-y^{1} = x-y$,which is clearly divisible by $(x-y)$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in \mathbb{N}$,i.e.,$x^{k}-y^{k} = m(x-y)$ for some integer $m$. (Equation $i$)
Step $3$: For $n=k+1$,we need to show $P(k+1): x^{k+1}-y^{k+1}$ is divisible by $(x-y)$.
$x^{k+1}-y^{k+1} = x^{k+1} - x^{k}y + x^{k}y - y^{k+1}$
$= x^{k}(x-y) + y(x^{k}-y^{k})$
Substituting from Equation $i$:
$= x^{k}(x-y) + y(m(x-y))$
$= (x-y)(x^{k} + my)$
Since $(x^{k} + my)$ is an integer,$x^{k+1}-y^{k+1}$ is divisible by $(x-y)$.
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 $P(n): n^{3}-n$ be divisible by $6$ for all $n \geq 2$.
Step $1$: For $n=2$,
$P(2): 2^{3}-2 = 8-2 = 6$,which is divisible by $6$.
Thus,$P(2)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \geq 2$,i.e.,$k^{3}-k = 6m$ for some integer $m$.
Step $3$: To prove $P(k+1)$ is true,we must show $(k+1)^{3}-(k+1)$ is divisible by $6$.
$(k+1)^{3}-(k+1) = (k^{3}+3k^{2}+3k+1) - k - 1$
$= (k^{3}-k) + 3k^{2}+3k$
$= (k^{3}-k) + 3k(k+1)$
Since $k^{3}-k = 6m$ and $k(k+1)$ is the product of two consecutive integers,it is divisible by $2$. Thus,$3k(k+1)$ is divisible by $3 \times 2 = 6$.
Therefore,$(k^{3}-k) + 3k(k+1) = 6m + 6n = 6(m+n)$,which is divisible by $6$.
Hence,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \geq 2$.

(A) Let $P(n): n(n^{2}+5)$ is divisible by $6$ for all $n \in N$.
Step $1$: For $n=1$,$P(1) = 1(1^{2}+5) = 6$,which is divisible by $6$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$k(k^{2}+5) = 6m$ for some integer $m$. $(i)$
Step $3$: We need to prove $P(k+1)$ is true,i.e.,$(k+1)((k+1)^{2}+5)$ is divisible by $6$.
Consider $(k+1)((k+1)^{2}+5) = (k+1)(k^{2}+2k+1+5) = (k+1)(k^{2}+2k+6)$
$= k(k^{2}+5) + k(2k+1) + 1(k^{2}+2k+6)$
$= k(k^{2}+5) + 2k^{2} + k + k^{2} + 2k + 6$
$= k(k^{2}+5) + 3k^{2} + 3k + 6$
$= 6m + 3k(k+1) + 6$
Since $k(k+1)$ is the product of two consecutive integers,it is always even,i.e.,$k(k+1) = 2p$ for some integer $p$.
$= 6m + 3(2p) + 6 = 6m + 6p + 6 = 6(m+p+1)$.
This is clearly divisible by $6$. 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 $P(n): n^{2} < 2^{n}$ for $n \geq 5, n \in \mathbb{N}$.
Step $1$: For $n = 5$,$P(5): 5^{2} < 2^{5} \implies 25 < 32$,which is true.
Step $2$: Assume $P(k)$ is true for some $k \geq 5$,i.e.,$k^{2} < 2^{k}$.
We need to show $P(k+1): (k+1)^{2} < 2^{k+1}$ is true.
Consider $(k+1)^{2} = k^{2} + 2k + 1$.
Since $k^{2} < 2^{k}$,we have $(k+1)^{2} < 2^{k} + 2k + 1$.
For $k \geq 5$,it can be shown that $2k + 1 < k^{2} < 2^{k}$.
Thus,$(k+1)^{2} < 2^{k} + 2^{k} = 2 \cdot 2^{k} = 2^{k+1}$.
Therefore,$P(k+1)$ is true whenever $P(k)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n \geq 5$.

(N/A) To prove: $P(n): \frac{d}{dx}(x^n) = nx^{n-1}$ for all positive integers $n$.
Step $1$: For $n=1$,
$P(1): \frac{d}{dx}(x) = 1 = 1 \cdot x^{1-1}$.
Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$.
That is,$P(k): \frac{d}{dx}(x^k) = kx^{k-1}$.
Step $3$: We need to prove $P(k+1)$ is true.
Consider $\frac{d}{dx}(x^{k+1}) = \frac{d}{dx}(x \cdot x^k)$.
Using the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.
$= x^k \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(x^k)$
$= x^k \cdot 1 + x \cdot (kx^{k-1})$
$= x^k + kx^k$
$= (k+1)x^k$
$= (k+1)x^{(k+1)-1}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Therefore,by the principle of mathematical induction,the statement $P(n)$ is true for every positive integer $n$.

(N/A) Let $P(n): 2n < (n+2)!$ for all $n \in \mathbb{N}$.
Step $1$: For $n=1$,$P(1): 2(1) < (1+2)! \implies 2 < 3! \implies 2 < 6$,which is true.
Step $2$: Assume $P(k)$ is true for some $k \in \mathbb{N}$,i.e.,$2k < (k+2)!$.
Step $3$: We need to prove $P(k+1): 2(k+1) < (k+3)!$.
Starting from the assumption $2k < (k+2)!$,we add $2$ to both sides:
$2k + 2 < (k+2)! + 2$
$2(k+1) < (k+2)! + 2$
Since $(k+2)! + 2 < (k+2)! \times (k+3)$ for all $k \ge 1$ (because $(k+2)! \times (k+3) - (k+2)! = (k+2)!(k+3-1) = (k+2)!(k+2) \ge 2! \times 2 = 4 > 2$),
we have $2(k+1) < (k+2)! + 2 < (k+3)!$.
Thus,$2(k+1) < (k+3)!$,which means $P(k+1)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.

(N/A) Let $P(n): \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{n}}$ for $n \geq 2$.
For $n=2$,$P(2): \sqrt{2} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} = 1 + \frac{1}{1.414} \approx 1 + 0.707 = 1.707$. Since $\sqrt{2} \approx 1.414$,$1.414 < 1.707$ is true.
Assume $P(k)$ is true for some $k \geq 2$: $\sqrt{k} < \sum_{i=1}^{k} \frac{1}{\sqrt{i}}$.
We need to show $P(k+1): \sqrt{k+1} < \sum_{i=1}^{k+1} \frac{1}{\sqrt{i}}$.
We know $\sqrt{k+1} - \sqrt{k} = \frac{(\sqrt{k+1} - \sqrt{k})(\sqrt{k+1} + \sqrt{k})}{\sqrt{k+1} + \sqrt{k}} = \frac{1}{\sqrt{k+1} + \sqrt{k}}$.
Since $\sqrt{k+1} + \sqrt{k} > 2\sqrt{k} > \sqrt{k+1}$,we have $\frac{1}{\sqrt{k+1} + \sqrt{k}} < \frac{1}{\sqrt{k+1}}$.
Thus,$\sqrt{k+1} < \sqrt{k} + \frac{1}{\sqrt{k+1}} < \sum_{i=1}^{k} \frac{1}{\sqrt{i}} + \frac{1}{\sqrt{k+1}} = \sum_{i=1}^{k+1} \frac{1}{\sqrt{i}}$.
Thus,$P(k+1)$ is true. By the Principle of Mathematical Induction,$P(n)$ is true for all $n \geq 2$.

(N/A) Let $P(n): 2+4+6+\ldots+2n = n^2+n$.
Step $1$: For $n=1$,$L.H.S. = 2$ and $R.H.S. = (1)^2+1 = 2$.
Since $L.H.S. = R.H.S.$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k$,i.e.,$2+4+6+\ldots+2k = k^2+k$.
Step $3$: For $n=k+1$,we need to show $P(k+1): 2+4+6+\ldots+2k+2(k+1) = (k+1)^2+(k+1)$.
$L.H.S. = (2+4+6+\ldots+2k) + 2(k+1)$
$= (k^2+k) + 2k+2$
$= k^2+3k+2$
$= (k^2+2k+1) + (k+1)$
$= (k+1)^2 + (k+1) = R.H.S.$
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) $P(n): 1+2+2^{2}+\ldots+2^{n}=2^{n+1}-1$
Step $1$: For $n=1$,
$L.H.S. = 1+2 = 3$
$R.H.S. = 2^{1+1}-1 = 2^{2}-1 = 4-1 = 3$
Since $L.H.S. = R.H.S.$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k$,i.e.,
$1+2+2^{2}+\ldots+2^{k}=2^{k+1}-1 \quad \dots(i)$
Step $3$: We need to show $P(k+1)$ is true.
$P(k+1): 1+2+2^{2}+\ldots+2^{k}+2^{k+1} = 2^{(k+1)+1}-1$
$L.H.S. = (1+2+2^{2}+\ldots+2^{k}) + 2^{k+1}$
Using equation $(i)$:
$L.H.S. = (2^{k+1}-1) + 2^{k+1}$
$= 2 \times 2^{k+1} - 1$
$= 2^{k+2} - 1$
$= 2^{(k+1)+1} - 1 = R.H.S.$
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) $P(n): 1+5+9+\ldots+(4 n-3)=n(2 n-1)$
For $n=1, \quad L.H.S.=1$
$R$.$H$.$S$. $=1(2(1)-1) = 1(1) = 1$
$\therefore L.H.S. = R.H.S.$
$\therefore P(1)$ is true.
Assume that $P(k)$ is true for some $k \in \mathbb{N}$.
$P(k): 1+5+9+\ldots+(4 k-3)=k(2 k-1) \quad \ldots(i)$
For $n=k+1$,we need to show $P(k+1)$ is true:
$L.H.S. = [1+5+9+\ldots+(4 k-3)] + (4(k+1)-3)$
$= k(2 k-1) + (4 k+4-3) \quad (\text{using } (i))$
$= 2 k^{2}-k+4 k+1$
$= 2 k^{2}+3 k+1$
$= 2 k^{2}+2 k+k+1$
$= 2k(k+1) + 1(k+1)$
$= (k+1)(2 k+1)$
$= (k+1)[2(k+1)-1] = R.H.S.$
$\therefore P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.

(A) Let $P(n)$ be the statement $a_{n} = 3 \cdot 7^{n-1}$ for $n \in N$.
Step $1$: For $n=1$,$a_{1} = 3 \cdot 7^{1-1} = 3 \cdot 7^{0} = 3(1) = 3$. This matches the given $a_{1} = 3$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$a_{k} = 3 \cdot 7^{k-1}$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$a_{k+1} = 3 \cdot 7^{(k+1)-1} = 3 \cdot 7^{k}$.
From the given recurrence relation,$a_{k+1} = 7 a_{k}$.
Substituting the assumption $a_{k} = 3 \cdot 7^{k-1}$,we get:
$a_{k+1} = 7 \cdot (3 \cdot 7^{k-1}) = 3 \cdot 7^{1} \cdot 7^{k-1} = 3 \cdot 7^{k}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
By the Principle of Mathematical Induction,$a_{n} = 3 \cdot 7^{n-1}$ is true for all $n \in N$.

(N/A) Let $P(n): b_{n}=5+4n$ for all $n \in \mathbb{N}$.
Step $1$: Base case for $n=1$.
$P(1): b_{1}=5+4(1)=9$.
From the recurrence relation $b_{k}=4+b_{k-1}$,for $k=1$,$b_{1}=4+b_{0}=4+5=9$.
Since $L.H.S. = R.H.S.$,$P(1)$ is true.
Step $2$: Inductive hypothesis.
Assume $P(k)$ is true for some $k \in \mathbb{N}$,i.e.,$b_{k}=5+4k$.
Step $3$: Inductive step.
We need to show $P(k+1)$ is true,i.e.,$b_{k+1}=5+4(k+1)$.
$b_{k+1}=4+b_{k}$ (by definition).
Substituting the hypothesis: $b_{k+1}=4+(5+4k) = 5+4(k+1)$.
Thus,$P(k+1)$ is true.
Conclusion: By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.

(N/A) Let $P(n)$ be the statement $d_{n} = \frac{2}{n!}$ for $n \in N$.
Step $1$: For $n=1$,$d_{1} = \frac{2}{1!} = \frac{2}{1} = 2$. This matches the given $d_{1}=2$. So,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$d_{k} = \frac{2}{k!}$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$d_{k+1} = \frac{2}{(k+1)!}$.
Given $d_{k+1} = \frac{d_{k}}{k+1}$.
Substituting the assumption $d_{k} = \frac{2}{k!}$,we get:
$d_{k+1} = \frac{2/k!}{k+1} = \frac{2}{k!(k+1)} = \frac{2}{(k+1)!}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.

(A) Let $P(n): \cos \alpha + \cos (\alpha + \beta) + \ldots + \cos [\alpha + (n-1)\beta] = \frac{\cos \left[\alpha + \left(\frac{n-1}{2}\right) \beta\right] \sin \left(\frac{n\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$
For $n=1$,$L.H.S. = \cos \alpha$
$R.H.S. = \frac{\cos \left[\alpha + \left(\frac{1-1}{2}\right) \beta\right] \sin \left(\frac{\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} = \cos \alpha$
Since $L.H.S. = R.H.S.$,$P(1)$ is true.
Assume $P(k)$ is true for some $k \in N$:
$P(k): \cos \alpha + \cos (\alpha + \beta) + \ldots + \cos [\alpha + (k-1)\beta] = \frac{\cos \left[\alpha + \left(\frac{k-1}{2}\right) \beta\right] \sin \left(\frac{k\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \quad \ldots (i)$
For $n=k+1$,we need to prove:
$P(k+1): \sum_{i=0}^{k} \cos(\alpha + i\beta) = \frac{\cos \left[\alpha + \frac{k\beta}{2}\right] \sin \left(\frac{(k+1)\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$
$L.H.S. = [\cos \alpha + \ldots + \cos(\alpha + (k-1)\beta)] + \cos(\alpha + k\beta)$
$= \frac{\cos \left[\alpha + \frac{(k-1)\beta}{2}\right] \sin \left(\frac{k\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} + \cos(\alpha + k\beta)$
$= \frac{\cos \left(\alpha + \frac{k\beta}{2} - \frac{\beta}{2}\right) \sin \left(\frac{k\beta}{2}\right) + \sin \left(\frac{\beta}{2}\right) \cos(\alpha + k\beta)}{\sin \left(\frac{\beta}{2}\right)}$
Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$= \frac{[\cos(\alpha + \frac{k\beta}{2}) \cos \frac{\beta}{2} + \sin(\alpha + \frac{k\beta}{2}) \sin \frac{\beta}{2}] \sin \frac{k\beta}{2} + \sin \frac{\beta}{2} \cos(\alpha + k\beta)}{\sin \frac{\beta}{2}}$
After simplification using trigonometric identities,we get:
$= \frac{\cos(\alpha + \frac{k\beta}{2}) [\sin \frac{k\beta}{2} \cos \frac{\beta}{2} + \cos \frac{k\beta}{2} \sin \frac{\beta}{2}]}{\sin \frac{\beta}{2}}$
$= \frac{\cos(\alpha + \frac{k\beta}{2}) \sin(\frac{k\beta + \beta}{2})}{\sin \frac{\beta}{2}} = \frac{\cos(\alpha + \frac{k\beta}{2}) \sin(\frac{(k+1)\beta}{2})}{\sin \frac{\beta}{2}}$
Thus,$P(k+1)$ is true. By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.

(A) $P(n): \cos \theta \cdot \cos 2 \theta \cdot \cos 2^{2} \theta \ldots \cos 2^{n-1} \theta = \frac{\sin 2^{n} \theta}{2^{n} \sin \theta}, \quad \forall n \in N$
For $n=1$,$L.H.S. = \cos \theta$
$R.H.S. = \frac{\sin 2 \theta}{2 \sin \theta} = \frac{2 \sin \theta \cos \theta}{2 \sin \theta} = \cos \theta$
$\therefore L.H.S. = R.H.S.$
$\therefore P(1)$ is true.
Assume that $P(k)$ is true for some $k \in N$,i.e.,$\cos \theta \cdot \cos 2 \theta \cdot \cos 2^{2} \theta \ldots \cos 2^{k-1} \theta = \frac{\sin 2^{k} \theta}{2^{k} \sin \theta} \quad (i)$
For $n=k+1$,we need to prove $P(k+1): \cos \theta \cdot \cos 2 \theta \ldots \cos 2^{k-1} \theta \cdot \cos 2^{k} \theta = \frac{\sin 2^{k+1} \theta}{2^{k+1} \sin \theta}$
$L.H.S. = \left( \frac{\sin 2^{k} \theta}{2^{k} \sin \theta} \right) \cdot \cos 2^{k} \theta \quad (\text{using } (i))$
$= \frac{2 \sin 2^{k} \theta \cos 2^{k} \theta}{2 \cdot 2^{k} \sin \theta} = \frac{\sin 2(2^{k} \theta)}{2^{k+1} \sin \theta} = \frac{\sin 2^{k+1} \theta}{2^{k+1} \sin \theta}$
$\therefore P(k+1)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.

(N/A) $P(n): \sin \theta + \sin 2\theta + \ldots + \sin n\theta = \frac{\sin \frac{n\theta}{2} \sin \frac{(n+1)\theta}{2}}{\sin \frac{\theta}{2}}, n \in N$
For $n=1$,$L.H.S. = \sin \theta$.
$R.H.S. = \frac{\sin \frac{\theta}{2} \sin \frac{2\theta}{2}}{\sin \frac{\theta}{2}} = \sin \theta$.
Since $L.H.S. = R.H.S.$,$P(1)$ is true.
Assume $P(k)$ is true for some $k \in N$:
$\sin \theta + \sin 2\theta + \ldots + \sin k\theta = \frac{\sin \frac{k\theta}{2} \sin \frac{(k+1)\theta}{2}}{\sin \frac{\theta}{2}}$.
For $n=k+1$,$L.H.S. = (\sin \theta + \ldots + \sin k\theta) + \sin(k+1)\theta$
$= \frac{\sin \frac{k\theta}{2} \sin \frac{(k+1)\theta}{2}}{\sin \frac{\theta}{2}} + \sin(k+1)\theta$
$= \frac{\sin \frac{k\theta}{2} \sin \frac{(k+1)\theta}{2} + \sin(k+1)\theta \sin \frac{\theta}{2}}{\sin \frac{\theta}{2}}$
Using $2\sin A \sin B = \cos(A-B) - \cos(A+B)$:
$= \frac{\frac{1}{2} [\cos(\frac{k\theta}{2} - \frac{(k+1)\theta}{2}) - \cos(\frac{k\theta}{2} + \frac{(k+1)\theta}{2})] + \frac{1}{2} [\cos((k+1)\theta - \frac{\theta}{2}) - \cos((k+1)\theta + \frac{\theta}{2})]}{\sin \frac{\theta}{2}}$
$= \frac{\cos \frac{\theta}{2} - \cos \frac{(2k+1)\theta}{2} + \cos \frac{(2k+1)\theta}{2} - \cos \frac{(2k+3)\theta}{2}}{2 \sin \frac{\theta}{2}}$
$= \frac{\cos \frac{\theta}{2} - \cos \frac{(2k+3)\theta}{2}}{2 \sin \frac{\theta}{2}}$
Using $\cos C - \cos D = 2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$:
$= \frac{2 \sin \frac{(k+2)\theta}{2} \sin \frac{(k+1)\theta}{2}}{2 \sin \frac{\theta}{2}} = \frac{\sin \frac{(k+1)\theta}{2} \sin \frac{(k+2)\theta}{2}}{\sin \frac{\theta}{2}} = R.H.S.$
Thus,$P(k+1)$ is true. By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.

(N/A) $P(n): \frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$ is a natural number,$n \in N$.
For $n=1$,$P(1) = \frac{1^{5}}{5} + \frac{1^{3}}{3} + \frac{7(1)}{15} = \frac{3+5+7}{15} = \frac{15}{15} = 1$,which is a natural number.
Therefore,$P(1)$ is true.
Assume that $P(k)$ is true for some $k \in N$,i.e.,$\frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7k}{15} = m$,where $m \in N$.
For $n=k+1$,we have $P(k+1) = \frac{(k+1)^{5}}{5} + \frac{(k+1)^{3}}{3} + \frac{7(k+1)}{15}$.
Expanding the terms: $P(k+1) = \frac{k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1}{5} + \frac{k^{3}+3k^{2}+3k+1}{3} + \frac{7k+7}{15}$.
Rearranging: $P(k+1) = (\frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7k}{15}) + (k^{4} + 2k^{3} + 2k^{2} + k) + (k^{2} + k) + (\frac{1}{5} + \frac{1}{3} + \frac{7}{15})$.
$P(k+1) = m + k^{4} + 2k^{3} + 3k^{2} + 2k + (\frac{3+5+7}{15}) = m + k^{4} + 2k^{3} + 3k^{2} + 2k + 1$.
Since $m, k \in N$,$P(k+1)$ is also a natural number.
Thus,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.

(A) Let $P(n): \frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} > \frac{13}{24}$.
Step $1$: For $n = 2$,the expression is $\frac{1}{2+1} + \frac{1}{2+2} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12} = \frac{14}{24}$.
Since $\frac{14}{24} > \frac{13}{24}$,$P(2)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N, k > 1$,i.e.,$\frac{1}{k+1} + \frac{1}{k+2} + \ldots + \frac{1}{2k} > \frac{13}{24}$.
Step $3$: For $n = k+1$,we need to show $P(k+1) > \frac{13}{24}$.
$P(k+1) = \frac{1}{k+2} + \frac{1}{k+3} + \ldots + \frac{1}{2k} + \frac{1}{2k+1} + \frac{1}{2k+2}$.
We can write $P(k+1) = P(k) - \frac{1}{k+1} + \frac{1}{2k+1} + \frac{1}{2k+2}$.
$P(k+1) = P(k) + \frac{1}{2k+1} + \frac{1}{2(k+1)} - \frac{1}{k+1} = P(k) + \frac{1}{2k+1} - \frac{1}{2(k+1)}$.
Since $2k+2 > 2k+1$,it follows that $\frac{1}{2k+1} > \frac{1}{2k+2}$,so $\frac{1}{2k+1} - \frac{1}{2(k+1)} > 0$.
Thus,$P(k+1) > P(k) > \frac{13}{24}$.
Therefore,$P(k+1)$ is true whenever $P(k)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n > 1$.

(N/A) $P(n):$ The number of subsets of a set having $n$ elements is $2^{n}$,where $n \in N$.
For $n=1$:
Let $A$ be a set with one element,$A = \{x\}$.
The subsets of $A$ are $\phi$ and $A$.
The number of subsets of $A$ is $2 = 2^{1}$.
Thus,$P(1)$ is true.
Assume that $P(k)$ is true for some $k \in N$,i.e.,a set with $k$ elements has $2^{k}$ subsets.
Now,we prove for $n = k+1$.
Let $A = \{a_{1}, a_{2}, \ldots, a_{k}, a_{k+1}\}$.
The subsets of $A$ can be divided into two types: those that do not contain $a_{k+1}$ and those that do contain $a_{k+1}$.
The number of subsets not containing $a_{k+1}$ is the number of subsets of $\{a_{1}, a_{2}, \ldots, a_{k}\}$,which is $2^{k}$ by the assumption $P(k)$.
The number of subsets containing $a_{k+1}$ is also $2^{k}$ (each subset is formed by adding $a_{k+1}$ to each of the $2^{k}$ subsets of $\{a_{1}, a_{2}, \ldots, a_{k}\}$).
Therefore,the total number of subsets of $A$ is $2^{k} + 2^{k} = 2 \cdot 2^{k} = 2^{k+1}$.
Thus,$P(k+1)$ is true.
Hence,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.

(B) Given the statement $P(n): n^2 - n + 37$ is prime.
For $n = 3$,$P(3) = 3^2 - 3 + 37 = 9 - 3 + 37 = 43$. Since $43$ is a prime number,$P(3)$ is true.
For $n = 5$,$P(5) = 5^2 - 5 + 37 = 25 - 5 + 37 = 57$. Since $57 = 3 \times 19$,it is not a prime number,so $P(5)$ is false.
Therefore,$P(3)$ is true but $P(5)$ is false.

(C) We are given the statement $P(n): 2^{n} < n!$.
We test for positive integer values of $n$:
For $n=1$: $2^{1} = 2$ and $1! = 1$. Since $2 < 1$ is false,$P(1)$ is false.
For $n=2$: $2^{2} = 4$ and $2! = 2$. Since $4 < 2$ is false,$P(2)$ is false.
For $n=3$: $2^{3} = 8$ and $3! = 6$. Since $8 < 6$ is false,$P(3)$ is false.
For $n=4$: $2^{4} = 16$ and $4! = 24$. Since $16 < 24$ is true,$P(4)$ is true.
Therefore,the smallest positive integer for which $P(n)$ is true is $n=4$.

(D) Given the statement $P(n): 2^{n} < n!$.
We test for positive integers $n = 1, 2, 3, 4, \dots$
For $n = 1$: $2^{1} < 1! \implies 2 < 1$,which is false.
For $n = 2$: $2^{2} < 2! \implies 4 < 2$,which is false.
For $n = 3$: $2^{3} < 3! \implies 8 < 6$,which is false.
For $n = 4$: $2^{4} < 4! \implies 16 < 24$,which is true.
Thus,the smallest positive integer for which $P(n)$ is true is $4$.

(D) Let $P(n) = n^{3} + 2n$.
For $n = 1$,$P(1) = 1^{3} + 2(1) = 1 + 2 = 3$.
For $n = 2$,$P(2) = 2^{3} + 2(2) = 8 + 4 = 12$.
For $n = 3$,$P(3) = 3^{3} + 2(3) = 27 + 6 = 33$.
We observe that $3, 12, 33$ are all divisible by $3$.
Alternatively,$n^{3} + 2n = n^{3} - n + 3n = n(n^{2} - 1) + 3n = (n-1)n(n+1) + 3n$.
Since $(n-1)n(n+1)$ is the product of three consecutive integers,it is always divisible by $3! = 6$.
However,the expression $n^{3} + 2n$ is always divisible by $3$ for any positive integer $n$.

(C) Let $p(n) = 2 \cdot 4^{2n+1} + 3^{3n+1}$.
For $n = 1$,$p(1) = 2 \cdot 4^{2(1)+1} + 3^{3(1)+1} = 2 \cdot 4^3 + 3^4 = 2 \cdot 64 + 81 = 128 + 81 = 209$.
Since $209 = 11 \times 19$,$p(1)$ is divisible by $11$.
Assume $p(k) = 2 \cdot 4^{2k+1} + 3^{3k+1}$ is divisible by $11$ for some positive integer $k$.
Now,consider $p(k+1) = 2 \cdot 4^{2(k+1)+1} + 3^{3(k+1)+1} = 2 \cdot 4^{2k+3} + 3^{3k+4}$.
$p(k+1) = 2 \cdot 4^{2k+1} \cdot 4^2 + 3^{3k+1} \cdot 3^3 = 16 \cdot (2 \cdot 4^{2k+1}) + 27 \cdot 3^{3k+1}$.
$p(k+1) = 16 \cdot (2 \cdot 4^{2k+1} + 3^{3k+1}) + (27 - 16) \cdot 3^{3k+1}$.
$p(k+1) = 16 \cdot p(k) + 11 \cdot 3^{3k+1}$.
Since $p(k)$ is divisible by $11$ and $11 \cdot 3^{3k+1}$ is clearly divisible by $11$,$p(k+1)$ is divisible by $11$.
By the principle of mathematical induction,$2 \cdot 4^{2n+1} + 3^{3n+1}$ is divisible by $11$ for all positive integers $n$.

(C) Given the inequality $8n + 16 \leq 2^n$.
We can rewrite the expression as $2^3(n + 2) \leq 2^n$.
Let us test the options:
For $n = 5$: $8(5) + 16 = 40 + 16 = 56$ and $2^5 = 32$. Since $56 \not\leq 32$,this is false.
For $n = 6$: $8(6) + 16 = 48 + 16 = 64$ and $2^6 = 64$. Since $64 \leq 64$ is true,the statement holds for $n = 6$.

(C) We know that the sum of the first $n$ squares is given by $1^2+2^2+3^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3+3n^2+n}{6}$.
Given $P(n): 1^2+2^2+3^2+\ldots+n^2 = \frac{6(n-1)(n-2) \ldots(n-2020) + 2n^3+3n^2+n}{6}$.
Substituting the standard sum formula into the given equation,we get:
$\frac{n(n+1)(2n+1)}{6} = (n-1)(n-2) \ldots(n-2020) + \frac{n(n+1)(2n+1)}{6}$.
This equality holds if and only if $(n-1)(n-2) \ldots(n-2020) = 0$.
This product is zero when $n \in \{1, 2, 3, \ldots, 2020\}$.
Therefore,$P(n)$ is true for all $n \leq 2020$.

(B) Given the inequality: $2^n > n+1$.
For $n=1$: $2^1 > 1+1 \Rightarrow 2 > 2$,which is false.
For $n=2$: $2^2 > 2+1 \Rightarrow 4 > 3$,which is true.
For $n=3$: $2^3 > 3+1 \Rightarrow 8 > 4$,which is true.
Using the principle of mathematical induction,assume $2^k > k+1$ for some $k \geq 2$.
We want to show $2^{k+1} > (k+1)+1 = k+2$.
Since $2^k > k+1$,multiplying both sides by $2$ gives $2^{k+1} > 2k+2$.
We know $2k+2 = (k+2) + k$.
Since $k \geq 2$,$k > 0$,therefore $2k+2 > k+2$.
Thus,$2^{k+1} > k+2$ is true for all $n \geq 2$.

(B) Given the recurrence relation: $a_0=1$ and $a_{n+1}=3n^2+n+a_n$.
We calculate the first few terms:
For $n=0$: $a_1 = 3(0)^2+0+a_0 = 1$.
For $n=1$: $a_2 = 3(1)^2+1+a_1 = 3+1+1 = 5$.
For $n=2$: $a_3 = 3(2)^2+2+a_2 = 12+2+5 = 19$.
Now,we test the options for $n=3$:
Option $A$: $3^3+3^2+1 = 27+9+1 = 37$.
Option $B$: $3^3-3^2+1 = 27-9+1 = 19$.
Option $C$: $3^3-3^2 = 27-9 = 18$.
Option $D$: $3^3+3^2 = 27+9 = 36$.
Since $a_3=19$,Option $B$ is the correct formula.

(A) Given,$P(n): 2+2^2+2^3+\ldots+2^n = 2^{n+1}-2$.
Assume $P(m)$ is true,so $2+2^2+\ldots+2^m = 2^{m+1}-2$.
Now,consider $P(m+1)$:
$P(m+1) = (2+2^2+\ldots+2^m) + 2^{m+1}$
$= (2^{m+1}-2) + 2^{m+1}$
$= 2 \cdot 2^{m+1} - 2 = 2^{m+2}-2$.
Since $P(m+1)$ holds true whenever $P(m)$ is true,the implication $P(m) \Rightarrow P(m+1)$ is correct.

(B) Given,$a_0 = 1$ and $a_{n+1} = 3n^2 + n + a_n$.
For $n = 0$: $a_1 = 3(0)^2 + 0 + a_0 = 0 + 0 + 1 = 1$.
For $n = 1$: $a_2 = 3(1)^2 + 1 + a_1 = 3 + 1 + 1 = 5$.
Testing the options:
For option $(B)$,$P(n) = n^3 - n^2 + 1$:
$P(0) = 0^3 - 0^2 + 1 = 1 = a_0$.
$P(1) = 1^3 - 1^2 + 1 = 1 = a_1$.
$P(2) = 2^3 - 2^2 + 1 = 8 - 4 + 1 = 5 = a_2$.
Since the values match for the initial terms,the correct expression is $a_n = n^3 - n^2 + 1$.

(A) We know that the sum of the squares of the first $n$ natural numbers is given by the formula:
$S_n = 1^2+2^2+3^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6}$
Expanding this expression,we get:
$S_n = \frac{2n^3+3n^2+n}{6} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$
Since $n \in \mathbb{N}$,$n \ge 1$,therefore $\frac{n^2}{2} + \frac{n}{6} > 0$.
Thus,$S_n = \frac{n^3}{3} + (\text{positive terms}) > \frac{n^3}{3}$.
Comparing this with the given inequality $S_n > x$,we find that $x = \frac{n^3}{3}$.

(C) We check each statement for $n = 1$:
$(A)$ $(2(1) + 7) < (1 + 3)^2 \implies 9 < 16$,which is True.
$(B)$ $1^2 > \frac{1^3}{3} \implies 1 > \frac{1}{3}$,which is True.
$(C)$ For $n = 1$,$3 \cdot 5^{2(1) + 1} + 2^{3(1) + 1} = 3 \cdot 5^3 + 2^4 = 3 \cdot 125 + 16 = 375 + 16 = 391$.
Checking divisibility: $391 \div 23 = 17$. Since $391$ is divisible by $23$,this statement is True for $n=1$.
Wait,let us re-check $n=2$ for $(C)$: $3 \cdot 5^5 + 2^7 = 3 \cdot 3125 + 128 = 9375 + 128 = 9503$.
$9503 \div 23 = 413.17$. Thus,it is not divisible by $23$ for all $n$.
$(D)$ For $n = 1$,$2 = \frac{1(5(1) - 1)}{2} = \frac{4}{2} = 2$,which is True.
Therefore,the statement that is not true for all $n \in N$ is $(C)$.

(B) Let $P(n)$ be the statement $2^{2n+1} + 3^{2n+1}$.
For $n = 1$:
$P(1) = 2^{2(1)+1} + 3^{2(1)+1} = 2^3 + 3^3 = 8 + 27 = 35$.
Since $35$ is divisible by $5$,$P(1)$ is true.
Assume $P(m)$ is true for some $m \in N$,i.e.,$2^{2m+1} + 3^{2m+1} = 5k$ for some integer $k$.
Then $2^{2m+1} = 5k - 3^{2m+1} \quad (i)$.
Now,consider $P(m+1)$:
$P(m+1) = 2^{2(m+1)+1} + 3^{2(m+1)+1} = 2^{2m+3} + 3^{2m+3}$.
$P(m+1) = 2^2 \cdot 2^{2m+1} + 3^2 \cdot 3^{2m+1}$.
Substituting from $(i)$:
$P(m+1) = 4(5k - 3^{2m+1}) + 9 \cdot 3^{2m+1}$.
$P(m+1) = 20k - 4 \cdot 3^{2m+1} + 9 \cdot 3^{2m+1}$.
$P(m+1) = 20k + 5 \cdot 3^{2m+1} = 5(4k + 3^{2m+1})$.
Since $5(4k + 3^{2m+1})$ is divisible by $5$,$P(m+1)$ is true.
By the principle of mathematical induction,$2^{2n+1} + 3^{2n+1}$ is divisible by $5$ for all $n \in N$.

(B) Let $P(n): 4^n+15n-1$ be divisible by $9$.
For $n=1$,$P(1)=4^1+15(1)-1=18$,which is divisible by $9$.
Therefore,$P(1)$ is true.
Assume $P(k)$ is true for some $k \in N$. Then,$4^k+15k-1=9\lambda$ for some $\lambda \in N$.
We need to show that $P(k+1)$ is true,i.e.,$4^{k+1}+15(k+1)-1$ is divisible by $9$.
$4^{k+1}+15(k+1)-1 = 4 \cdot 4^k + 15k + 15 - 1$
Substitute $4^k = 9\lambda - 15k + 1$:
$= 4(9\lambda - 15k + 1) + 15k + 14$
$= 36\lambda - 60k + 4 + 15k + 14$
$= 36\lambda - 45k + 18$
$= 9(4\lambda - 5k + 2)$,which is divisible by $9$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true. By the Principle of Mathematical Induction,$4^n+15n-1$ is divisible by $9$ for all $n \in N$.

(A) Let $P(n) = 3(5^{2n+1}) + 2^{3n+1}$.
For $n = 1$,$P(1) = 3(5^3) + 2^4 = 3(125) + 16 = 375 + 16 = 391$.
Since $391 = 17 \times 23$,the expression is divisible by $17$.
Alternatively,we can write:
$3(5^{2n+1}) + 2^{3n+1} = 15(25^n) + 2(8^n)$
$= 15(25^n) - 15(8^n) + 15(8^n) + 2(8^n)$
$= 15(25^n - 8^n) + 17(8^n)$
Since $(25^n - 8^n)$ is divisible by $(25 - 8) = 17$,the entire expression is divisible by $17$.

(B) Given $P(n) = 3^{2n+1} + 2^{n+2}$.
For $n = 1$,$P(1) = 3^{2(1)+1} + 2^{1+2} = 3^3 + 2^3 = 27 + 8 = 35$.
Since $35 = 5 \times 7$,$P(1)$ is divisible by the prime integers $5$ and $7$.
Thus,there exists at least one prime integer that divides $P(n)$ for $n = 1$.
For any $n \in N$,$P(n) > 1$,and by the Fundamental Theorem of Arithmetic,every integer greater than $1$ has at least one prime divisor.
Therefore,there exists a prime integer which divides $P(n)$.