nth term of special series, Sum to n terms and Infinite number of terms Questions in English

(B) The $r^{th}$ term of the given series is $T_r = r[(r + 1) - \omega][(r + 1) - \omega^2]$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$ and $\omega^3 = 1$.
$T_r = r[(r + 1)^2 - (\omega + \omega^2)(r + 1) + \omega^3] = r[(r + 1)^2 + (r + 1) + 1] = r[r^2 + 2r + 1 + r + 1 + 1] = r(r^2 + 3r + 3) = r^3 + 3r^2 + 3r$.
The sum of the series is $S = \sum_{r=1}^{n-1} (r^3 + 3r^2 + 3r)$.
$S = \sum_{r=1}^{n-1} r^3 + 3 \sum_{r=1}^{n-1} r^2 + 3 \sum_{r=1}^{n-1} r$.
$S = [\frac{(n-1)n}{2}]^2 + 3 \cdot \frac{(n-1)n(2n-2+1)}{6} + 3 \cdot \frac{(n-1)n}{2}$.
$S = \frac{(n-1)^2 n^2}{4} + \frac{(n-1)n(2n-1)}{2} + \frac{3(n-1)n}{2}$.
$S = \frac{(n-1)n}{4} [(n-1)n + 2(2n-1) + 6] = \frac{(n-1)n}{4} [n^2 - n + 4n - 2 + 6] = \frac{1}{4}(n-1)n(n^2 + 3n + 4)$.

(C) The given series is $(1 \times 3) + (3 \times 5) + (5 \times 7) + (7 \times 9) + \dots$
Observe the first factors of each term: $1, 3, 5, 7, \dots$
This is an arithmetic progression with first term $a = 1$ and common difference $d = 2$.
The $n^{th}$ term of this sequence is $a_n = a + (n - 1)d = 1 + (n - 1)2 = 2n - 1$.
Observe the second factors of each term: $3, 5, 7, 9, \dots$
This is an arithmetic progression with first term $a = 3$ and common difference $d = 2$.
The $n^{th}$ term of this sequence is $b_n = 3 + (n - 1)2 = 2n + 1$.
Therefore,the $n^{th}$ term of the series is the product of the $n^{th}$ terms of these two sequences:
$T_n = (2n - 1)(2n + 1)$.

(D) The given series is $S = 1 - 2 + 3 - 4 + 5 - 6 + \dots + (-1)^{n-1}n$.
Case $I$: If $n$ is even,let $n = 2k$.
The sum is $(1-2) + (3-4) + \dots + ((2k-1) - 2k) = (-1) + (-1) + \dots + (-1)$ ($k$ times).
Sum $= -k = -\frac{n}{2}$.
Case $II$: If $n$ is odd,let $n = 2k+1$.
The sum is $(1-2) + (3-4) + \dots + ((2k-1) - 2k) + (2k+1) = -k + (2k+1) = k+1$.
Since $n = 2k+1$,we have $k = \frac{n-1}{2}$,so the sum is $\frac{n-1}{2} + 1 = \frac{n+1}{2}$.
Thus,the sum is $-\frac{n}{2}$ if $n$ is even and $\frac{n+1}{2}$ if $n$ is odd.

(A) The $n^{th}$ term is given by $T_n = \frac{n}{x} + y$.
To find the sum of $r$ terms,we calculate the sum $S_r = \sum_{n=1}^{r} T_n = \sum_{n=1}^{r} \left( \frac{n}{x} + y \right)$.
$S_r = \frac{1}{x} \sum_{n=1}^{r} n + \sum_{n=1}^{r} y$.
Using the formula for the sum of the first $r$ natural numbers,$\sum_{n=1}^{r} n = \frac{r(r + 1)}{2}$.
$S_r = \frac{1}{x} \left( \frac{r(r + 1)}{2} \right) + ry$.
$S_r = \left\{ \frac{r(r + 1)}{2x} \right\} + ry$.

(B) Given the sum of the first $n$ terms $S_n = 5n^2 + 2n$.
To find the second term $T_2$,we use the formula $T_2 = S_2 - S_1$.
First,calculate $S_1 = 5(1)^2 + 2(1) = 5 + 2 = 7$.
Next,calculate $S_2 = 5(2)^2 + 2(2) = 5(4) + 4 = 20 + 4 = 24$.
Therefore,$T_2 = S_2 - S_1 = 24 - 7 = 17$.

(B) The given series is $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$
The $n^{th}$ term $T_n$ can be written as the product of the $n^{th}$ term of the sequence $2, 4, 6, \dots$ and the $n^{th}$ term of the sequence $4, 6, 8, \dots$
The $n^{th}$ term of $2, 4, 6, \dots$ is $2n$.
The $n^{th}$ term of $4, 6, 8, \dots$ is $2(n + 1)$.
Thus,$T_n = 2n \times 2(n + 1) = 4n(n + 1)$.
For the $20^{th}$ term,substitute $n = 20$:
$T_{20} = 4 \times 20 \times (20 + 1) = 80 \times 21 = 1680$.

(B) Let $S_n = 6 + 66 + 666 + \dots$ up to $n$ terms.
$S_n = 6(1 + 11 + 111 + \dots + n \text{ terms})$
$S_n = \frac{6}{9}(9 + 99 + 999 + \dots + n \text{ terms})$
$S_n = \frac{2}{3}((10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1))$
$S_n = \frac{2}{3}((10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + n \text{ times}))$
Using the sum of a geometric progression formula $\frac{a(r^n - 1)}{r - 1}$:
$S_n = \frac{2}{3} \left( \frac{10(10^n - 1)}{10 - 1} - n \right)$
$S_n = \frac{2}{3} \left( \frac{10(10^n - 1)}{9} - n \right)$
$S_n = \frac{2}{3} \left( \frac{10^{n+1} - 10 - 9n}{9} \right)$
$S_n = \frac{2(10^{n+1} - 9n - 10)}{27}$.

(C) The $k$-th term of the series is $T_k = 1 + x + x^2 + \dots + x^{k-1} = \frac{1 - x^k}{1 - x}$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{1 - x^k}{1 - x}$.
$S_n = \frac{1}{1 - x} \sum_{k=1}^{n} (1 - x^k) = \frac{1}{1 - x} \left[ \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} x^k \right]$.
$S_n = \frac{1}{1 - x} \left[ n - \frac{x(1 - x^n)}{1 - x} \right]$.
$S_n = \frac{n(1 - x) - x(1 - x^n)}{(1 - x)^2}$.

(B) Let $S_n = 3 + 33 + 333 + \dots$ to $n$ terms.
$S_n = 3(1 + 11 + 111 + \dots \text{ to } n \text{ terms})$
$S_n = \frac{3}{9}(9 + 99 + 999 + \dots \text{ to } n \text{ terms})$
$S_n = \frac{1}{3}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$
$S_n = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots \text{ to } n \text{ terms})]$
Using the sum of a geometric progression formula $S = \frac{a(r^n - 1)}{r - 1}$:
$S_n = \frac{1}{3} \left[ \frac{10(10^n - 1)}{10 - 1} - n \right]$
$S_n = \frac{1}{3} \left[ \frac{10^{n+1} - 10}{9} - n \right]$
$S_n = \frac{1}{3} \left[ \frac{10^{n+1} - 10 - 9n}{9} \right]$
$S_n = \frac{1}{27}(10^{n+1} - 9n - 10)$.

(B) The given series is an infinite geometric progression with the first term $a = 3$ and common ratio $r = \alpha$.
The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$,where $|r| < 1$.
Given,$\frac{3}{1 - \alpha} = \frac{45}{8}$.
Dividing both sides by $3$,we get $\frac{1}{1 - \alpha} = \frac{15}{8}$.
Taking the reciprocal,$1 - \alpha = \frac{8}{15}$.
Therefore,$\alpha = 1 - \frac{8}{15} = \frac{15 - 8}{15} = \frac{7}{15}$.

(D) The sum of an infinite $G.P.$ is given by the formula $S_{\infty} = \frac{a}{1 - r}$.
This formula is valid only when the common ratio $r$ satisfies the condition $|r| < 1$,which means $-1 < r < 1$.

(A) Let the first term be $a$ and the common ratio be $r$. The sum of the infinite geometric series is given by $S = \frac{a}{1-r} = 3$,so $a = 3(1-r) \dots (i)$.
The series formed by squaring its terms is $a^2, a^2r^2, a^2r^4, \dots$,which is a geometric series with first term $a^2$ and common ratio $r^2$. Its sum is $\frac{a^2}{1-r^2} = 3$.
Substituting $a = 3(1-r)$ into the second equation: $\frac{9(1-r)^2}{1-r^2} = 3$.
Since $1-r^2 = (1-r)(1+r)$,we have $\frac{9(1-r)^2}{(1-r)(1+r)} = 3$,which simplifies to $\frac{3(1-r)}{1+r} = 1$.
Solving for $r$: $3 - 3r = 1 + r$,so $4r = 2$,which gives $r = \frac{1}{2}$.
Substituting $r = \frac{1}{2}$ into $(i)$: $a = 3(1 - \frac{1}{2}) = \frac{3}{2}$.
The series is $\frac{3}{2}, \frac{3}{2}(\frac{1}{2}), \frac{3}{2}(\frac{1}{2})^2, \dots$,which is $\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}, \dots$.

(D) For an infinite $G.P.$,the sum $S = \frac{a}{1-r} = 4$ and the second term $ar = \frac{3}{4}$.
From the first equation,$a = 4(1-r)$.
Substituting $a$ into the second equation: $4(1-r)r = \frac{3}{4}$.
$r(1-r) = \frac{3}{16} \implies r - r^2 = \frac{3}{16} \implies 16r^2 - 16r + 3 = 0$.
Factoring the quadratic: $(4r - 3)(4r - 1) = 0$.
This gives $r = \frac{3}{4}$ or $r = \frac{1}{4}$.
If $r = \frac{1}{4}$,then $a = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.
If $r = \frac{3}{4}$,then $a = 4(1 - \frac{3}{4}) = 4(\frac{1}{4}) = 1$.
The possible pairs $(a, r)$ are $(3, \frac{1}{4})$ and $(1, \frac{3}{4})$.
Comparing with the options,$(a, r) = (3, \frac{1}{4})$ is correct.

(A) The given geometric progression is $\frac{\sqrt{2} + 1}{\sqrt{2} - 1}, \frac{1}{\sqrt{2}(\sqrt{2} - 1)}, \frac{1}{2}, \dots$
The first term $a = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} = \frac{(\sqrt{2} + 1)^2}{2 - 1} = (\sqrt{2} + 1)^2 = 3 + 2\sqrt{2}$.
The common ratio $r = \frac{1}{\sqrt{2}(\sqrt{2} - 1)} \div \frac{\sqrt{2} + 1}{\sqrt{2} - 1} = \frac{1}{\sqrt{2}(\sqrt{2} + 1)} = \frac{1}{2 + \sqrt{2}}$.
The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.
$S = \frac{(\sqrt{2} + 1)^2}{1 - \frac{1}{\sqrt{2}(\sqrt{2} + 1)}} = \frac{(\sqrt{2} + 1)^2}{\frac{\sqrt{2}(\sqrt{2} + 1) - 1}{\sqrt{2}(\sqrt{2} + 1)}} = \frac{(\sqrt{2} + 1)^2 \cdot \sqrt{2}(\sqrt{2} + 1)}{2 + \sqrt{2} - 1} = \frac{\sqrt{2}(\sqrt{2} + 1)^3}{1 + \sqrt{2}} = \sqrt{2}(\sqrt{2} + 1)^2$.

(B) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r} = 20 \quad (i)$.
The sum of the squares of the terms is given by $\frac{a^2}{1-r^2} = 100 \quad (ii)$.
We can rewrite $(ii)$ as $\frac{a}{1-r} \times \frac{a}{1+r} = 100$.
Substituting $(i)$ into this equation: $20 \times \frac{a}{1+r} = 100$,which simplifies to $\frac{a}{1+r} = 5 \quad (iii)$.
From $(i)$,$a = 20(1-r)$. Substituting this into $(iii)$:
$\frac{20(1-r)}{1+r} = 5$
$4(1-r) = 1+r$
$4 - 4r = 1 + r$
$3 = 5r$
$r = 3/5$.

(D) Let $x = 0.037037037...$
This is a geometric series with the first term $a = \frac{37}{1000}$ and common ratio $r = \frac{1}{1000}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
Substituting the values: $S = \frac{37/1000}{1 - 1/1000} = \frac{37/1000}{999/1000} = \frac{37}{999}$.
Simplifying the fraction: $\frac{37}{999} = \frac{1}{27}$.
Since both options $B$ and $D$ are mathematically equivalent,and $\frac{37}{999}$ is the direct representation of the repeating decimal,the correct choice is $D$ (or $B$ as they are equal).

(C) The given series $9 - 3 + 1 - \frac{1}{3} + \dots$ is a geometric progression $(G.P.)$.
Here,the first term $a = 9$ and the common ratio $r = \frac{-3}{9} = -\frac{1}{3}$.
The sum to infinity of a $G.P.$ is given by the formula $S_{\infty} = \frac{a}{1 - r}$,provided $|r| < 1$.
Substituting the values,we get $S_{\infty} = \frac{9}{1 - (-1/3)} = \frac{9}{1 + 1/3} = \frac{9}{4/3} = \frac{9 \times 3}{4} = \frac{27}{4}$.

(C) Let the given expression be $P = (32)(32)^{1/6}(32)^{1/36} \dots \infty$.
Using the property of exponents $a^m \times a^n = a^{m+n}$,we have:
$P = (32)^{1 + \frac{1}{6} + \frac{1}{36} + \dots \infty}$.
The exponent is an infinite geometric series with the first term $a = 1$ and common ratio $r = \frac{1}{6}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$.
$S = \frac{1}{1 - 1/6} = \frac{1}{5/6} = \frac{6}{5}$.
Therefore,$P = (32)^{6/5}$.
Since $32 = 2^5$,we have $P = (2^5)^{6/5} = 2^{5 \times (6/5)} = 2^6$.
$P = 64$.

(D) Let $S = 1 + 3x + 6x^2 + 10x^3 + \dots \infty$
Multiply by $x$: $xS = x + 3x^2 + 6x^3 + \dots \infty$
Subtracting the two equations: $S(1 - x) = 1 + 2x + 3x^2 + 4x^3 + \dots \infty$
Multiply by $x$ again: $xS(1 - x) = x + 2x^2 + 3x^3 + \dots \infty$
Subtracting again: $S(1 - x) - xS(1 - x) = 1 + x + x^2 + x^3 + \dots \infty$
$S(1 - x)^2 = \frac{1}{1 - x}$
$S = \frac{1}{(1 - x)^3}$

(B) Let $T_n$ be the $n^{th}$ term and $S_n$ be the sum up to $n$ terms.
$S_n = 1 + 3 + 7 + 15 + 31 + \dots + T_n$
We can write the $n^{th}$ term as $T_n = 2^n - 1$.
Thus,$S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (2^k - 1)$.
$S_n = \sum_{k=1}^{n} 2^k - \sum_{k=1}^{n} 1$.
Using the sum formula for a geometric progression,$\sum_{k=1}^{n} 2^k = 2(2^n - 1) / (2 - 1) = 2^{n+1} - 2$.
Therefore,$S_n = (2^{n+1} - 2) - n = 2^{n+1} - n - 2$.

(B) Let the $n$-th term be $T_n$. The sequence is $2, 4, 7, 11, 16, \dots$
The differences between consecutive terms are $2, 3, 4, 5, \dots$,which form an arithmetic progression.
Thus,$T_n = T_1 + \sum_{k=1}^{n-1} (k+1) = 2 + \frac{(n-1)(2 + n)}{2} = 2 + \frac{n^2 + n - 2}{2} = \frac{4 + n^2 + n - 2}{2} = \frac{n^2 + n + 2}{2}$.
Now,the sum $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k^2 + k + 2}{2} = \frac{1}{2} \left[ \sum k^2 + \sum k + \sum 2 \right]$.
$S_n = \frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + 2n \right]$.
$S_n = \frac{n}{12} \left[ (n+1)(2n+1) + 3(n+1) + 12 \right] = \frac{n}{12} \left[ 2n^2 + 3n + 1 + 3n + 3 + 12 \right]$.
$S_n = \frac{n}{12} (2n^2 + 6n + 16) = \frac{n}{6} (n^2 + 3n + 8)$.

(C) Let the $n^{th}$ term be $T_n$. The series is $2, 4, 7, 11, \dots$
The differences between consecutive terms are $2, 3, 4, \dots$,which form an Arithmetic Progression.
Thus,$T_n = T_1 + \sum_{k=1}^{n-1} (k+1)$
$T_n = 2 + [2 + 3 + 4 + \dots + n]$
$T_n = 1 + [1 + 2 + 3 + \dots + n]$
Using the sum formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we get:
$T_n = 1 + \frac{n(n+1)}{2}$
$T_n = \frac{2 + n^2 + n}{2} = \frac{n^2 + n + 2}{2}$.

(C) The sum of the first $n$ terms is given by:
$S_n = \left(1 - \frac{1}{2}\right) + \left(1 - \frac{1}{2^2}\right) + \left(1 - \frac{1}{2^3}\right) + \dots + \left(1 - \frac{1}{2^n}\right)$
$S_n = \sum_{k=1}^{n} \left(1 - \frac{1}{2^k}\right) = n - \sum_{k=1}^{n} \left(\frac{1}{2}\right)^k$
Using the sum formula for a geometric progression $\sum_{k=1}^{n} ar^{k-1} = a\frac{1-r^n}{1-r}$,we have:
$S_n = n - \left[ \frac{1}{2} \left( \frac{1 - (1/2)^n}{1 - 1/2} \right) \right]$
$S_n = n - \left( 1 - \frac{1}{2^n} \right) = n - 1 + 2^{-n}$
Verification for $n=1$: $S_1 = 1 - 1 + 2^{-1} = 1/2$ (Correct).
Verification for $n=2$: $S_2 = 2 - 1 + 2^{-2} = 1 + 1/4 = 5/4$ (Correct).

(B) Let the given expression be $P = 2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \cdot 16^{1/32} \cdots$
Expressing all terms with base $2$:
$P = 2^{1/4} \cdot (2^2)^{1/8} \cdot (2^3)^{1/16} \cdot (2^4)^{1/32} \cdots$
$P = 2^{1/4 + 2/8 + 3/16 + 4/32 + \cdots} = 2^S$,where $S = \sum_{n=1}^{\infty} \frac{n}{2^{n+1}}$.
$S = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} + \cdots$ $(i)$
$\frac{1}{2}S = \frac{1}{8} + \frac{2}{16} + \frac{3}{32} + \cdots$ $(ii)$
Subtracting $(ii)$ from $(i)$:
$S - \frac{1}{2}S = \frac{1}{4} + (\frac{2}{8} - \frac{1}{8}) + (\frac{3}{16} - \frac{2}{16}) + \cdots$
$\frac{1}{2}S = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$
This is an infinite geometric series with $a = 1/4$ and $r = 1/2$.
$\frac{1}{2}S = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.
Thus,$S = 1$.
Therefore,$P = 2^1 = 2$.

(A) Let the $n^{th}$ term be $T_n$. The series is $2, 7, 14, 23, 34, \dots$
The differences between consecutive terms are $5, 7, 9, 11, \dots$,which form an Arithmetic Progression $(AP)$.
Let $T_n = an^2 + bn + c$.
For $n=1, T_1 = a + b + c = 2$
For $n=2, T_2 = 4a + 2b + c = 7$
For $n=3, T_3 = 9a + 3b + c = 14$
Subtracting equations: $(4a+2b+c) - (a+b+c) = 3a+b = 5$ and $(9a+3b+c) - (4a+2b+c) = 5a+b = 7$.
Subtracting these results: $2a = 2 \Rightarrow a = 1$.
Substituting $a=1$ into $3a+b=5$,we get $3+b=5 \Rightarrow b=2$.
Substituting $a=1, b=2$ into $a+b+c=2$,we get $1+2+c=2 \Rightarrow c=-1$.
Thus,$T_n = n^2 + 2n - 1$.
For $n=99$,$T_{99} = (99)^2 + 2(99) - 1 = 9801 + 198 - 1 = 9998$.

(A) The $n^{th}$ set $S_n$ contains $n$ terms.
The first term of $S_n$ is given by $a_n = 1 + \sum_{k=1}^{n-1} k = 1 + \frac{(n-1)n}{2}$.
For $n = 50$,the first term is $a_{50} = 1 + \frac{49 \times 50}{2} = 1 + 1225 = 1226$.
The set $S_{50}$ contains $50$ consecutive integers starting from $1226$,so $S_{50} = \{1226, 1227, \dots, 1275\}$.
The sum of these $50$ terms is an arithmetic progression sum: $Sum = \frac{n}{2}(2a + (n-1)d)$.
$Sum = \frac{50}{2}(2 \times 1226 + (50-1) \times 1) = 25(2452 + 49) = 25(2501) = 62525$.

(C) Let $T_k$ be the $k^{th}$ term of the series. The $k^{th}$ term is the sum of the first $k$ odd numbers,which is $T_k = k^2$.
The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$.
We need the sum of $(n - 1)$ terms,so we replace $n$ with $(n - 1)$ in the formula for $S_n$:
$S_{n-1} = \frac{(n - 1)((n - 1) + 1)(2(n - 1) + 1)}{6}$
$S_{n-1} = \frac{(n - 1)(n)(2n - 2 + 1)}{6}$
$S_{n-1} = \frac{(n - 1)n(2n - 1)}{6}$.

(B) The given series is $1^2, 2 \cdot 2^2, 3^2, 2 \cdot 4^2, 5^2, 2 \cdot 6^2, \dots$
When $n$ is even,the sum $S_n = \frac{n(n + 1)^2}{2}$.
When $n$ is odd,the $n^{th}$ term is $n^2$. The sum of the first $n$ terms is $S_n = S_{n-1} + a_n$.
Since $n$ is odd,$n-1$ is even. Using the given formula for $n-1$ terms:
$S_{n-1} = \frac{(n-1)((n-1) + 1)^2}{2} = \frac{(n-1)n^2}{2}$.
Thus,$S_n = \frac{(n-1)n^2}{2} + n^2 = n^2 \left( \frac{n-1}{2} + 1 \right) = n^2 \left( \frac{n-1+2}{2} \right) = \frac{n^2(n+1)}{2}$.
Verification: For $n=1$,$S_1 = 1^2 = 1$. Using formula $(b)$: $\frac{1^2(1+1)}{2} = 1$. Correct.

(B) The given series is $2^2 + 4^2 + 6^2 + \dots + (2n)^2$.
This can be written as $2^2(1^2 + 2^2 + 3^2 + \dots + n^2)$.
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$.
Thus,the sum is $4 \times \frac{n(n + 1)(2n + 1)}{6} = \frac{2n(n + 1)(2n + 1)}{3}$.

(D) The $n^{th}$ term of the series is given by $T_n = 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k(k + 1)}{2}$.
$S_n = \frac{1}{2} \left( \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \right)$.
Using the standard summation formulas $\sum k^2 = \frac{n(n + 1)(2n + 1)}{6}$ and $\sum k = \frac{n(n + 1)}{2}$:
$S_n = \frac{1}{2} \left( \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \right)$.
$S_n = \frac{n(n + 1)}{4} \left( \frac{2n + 1}{3} + 1 \right) = \frac{n(n + 1)}{4} \left( \frac{2n + 4}{3} \right)$.
$S_n = \frac{n(n + 1) \cdot 2(n + 2)}{12} = \frac{n(n + 1)(n + 2)}{6}$.

(A) Given the $n^{th}$ term is $T_n = n(n + 1) = n^2 + n$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k)$.
Using the standard summation formulas $\sum k^2 = \frac{n(n + 1)(2n + 1)}{6}$ and $\sum k = \frac{n(n + 1)}{2}$,we get:
$S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}$
$S_n = \frac{n(n + 1)}{2} \left( \frac{2n + 1}{3} + 1 \right)$
$S_n = \frac{n(n + 1)}{2} \left( \frac{2n + 1 + 3}{3} \right) = \frac{n(n + 1)(2n + 4)}{6}$
$S_n = \frac{n(n + 1) \cdot 2(n + 2)}{6} = \frac{n(n + 1)(n + 2)}{3}$.

(B) The given series is $S = 3 \cdot 6 + 4 \cdot 7 + 5 \cdot 8 + \dots$ up to $(n - 2)$ terms.
Let the general term be $T_k = (k + 2)(k + 5)$ for $k = 1, 2, \dots, (n - 2)$.
Alternatively,we can write the sum as $S = \sum_{k=1}^{n} k(k+3) - (1 \cdot 4 + 2 \cdot 5) = \sum_{k=1}^{n} (k^2 + 3k) - (4 + 10)$.
Using the summation formulas $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$:
$S = \frac{n(n+1)(2n+1)}{6} + 3 \cdot \frac{n(n+1)}{2} - 14$.
$S = \frac{2n^3 + 3n^2 + n + 9n^2 + 9n - 84}{6} = \frac{2n^3 + 12n^2 + 10n - 84}{6}$.
Thus,the correct option is $B$.

(A) We are given the sum $\sum\limits_{i = 1}^n (3i - 2)$.
Using the linearity property of summation,we can write:
$\sum\limits_{i = 1}^n (3i - 2) = 3\sum\limits_{i = 1}^n i - \sum\limits_{i = 1}^n 2$
Since $\sum\limits_{i = 1}^n i = \frac{n(n + 1)}{2}$ and $\sum\limits_{i = 1}^n 2 = 2n$,we have:
$= 3 \left[ \frac{n(n + 1)}{2} \right] - 2n$
$= \frac{3n^2 + 3n - 4n}{2}$
$= \frac{3n^2 - n}{2}$
$= \frac{n(3n - 1)}{2}$.

(B) The $n$-th term of the series is $T_n = n^2(n + 1) = n^3 + n^2$.
The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n k^3 + \sum_{k=1}^n k^2$.
Using the standard summation formulas,$S_n = \left[ \frac{n(n + 1)}{2} \right]^2 + \frac{n(n + 1)(2n + 1)}{6}$.
Factoring out $\frac{n(n + 1)}{2}$,we get $S_n = \frac{n(n + 1)}{2} \left[ \frac{n(n + 1)}{2} + \frac{2n + 1}{3} \right]$.
Simplifying the expression inside the brackets: $\frac{3n^2 + 3n + 4n + 2}{6} = \frac{3n^2 + 7n + 2}{6}$.
Thus,$S_n = \frac{n(n + 1)(3n^2 + 7n + 2)}{12}$.

(C) The $n^{th}$ term of the series is $T_n = n(n + 1)(n + 2)$.
Expanding this,we get $T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n$.
To find the sum $S_n$,we calculate $\sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^3 + 3k^2 + 2k)$.
Using standard summation formulas:
$S_n = \left[ \frac{n(n + 1)}{2} \right]^2 + 3 \left[ \frac{n(n + 1)(2n + 1)}{6} \right] + 2 \left[ \frac{n(n + 1)}{2} \right]$.
Simplifying the expression:
$S_n = \frac{n^2(n + 1)^2}{4} + \frac{n(n + 1)(2n + 1)}{2} + n(n + 1)$.
Factoring out $\frac{n(n + 1)}{4}$:
$S_n = \frac{n(n + 1)}{4} [n(n + 1) + 2(2n + 1) + 4]$.
$S_n = \frac{n(n + 1)}{4} [n^2 + n + 4n + 2 + 4] = \frac{n(n + 1)}{4} [n^2 + 5n + 6]$.
$S_n = \frac{n(n + 1)(n + 2)(n + 3)}{4}$.

(C) The sum of the cubes of the first $n$ natural numbers is given by the formula:
$S_n = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4}$
Here,$n = 15$.
Substituting the value of $n$:
$S_{15} = \frac{15^2 \times (15+1)^2}{4} = \frac{225 \times 16^2}{4} = \frac{225 \times 256}{4} = 225 \times 64 = 14400$.

(B) The sum of the squares of the first $n$ natural numbers is given by $\Sigma n^2 = \frac{n(n+1)(2n+1)}{6}$.
The sum of the first $n$ natural numbers is given by $\Sigma n = \frac{n(n+1)}{2}$.
According to the problem,$\Sigma n^2 = \Sigma n + 330$.
Substituting the formulas: $\frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{2} + 330$.
Rearranging the terms: $\frac{n(n+1)}{2} \left[ \frac{2n+1}{3} - 1 \right] = 330$.
Simplifying the expression inside the bracket: $\frac{n(n+1)}{2} \left[ \frac{2n+1-3}{3} \right] = 330$.
$\frac{n(n+1)}{2} \cdot \frac{2(n-1)}{3} = 330$.
$n(n+1)(n-1) = 990$.
$n(n^2-1) = 990$.
$n^3 - n - 990 = 0$.
Testing values,for $n=10$: $10^3 - 10 = 1000 - 10 = 990$. Thus,$n = 10$.

(A) The given series is $\frac{1}{1} + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$
The $n^{th}$ term of the series is given by:
$T_n = \frac{1 + 2 + 3 + \dots + n}{n}$
Using the formula for the sum of the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}$,we get:
$T_n = \frac{\frac{n(n + 1)}{2}}{n}$
Simplifying the expression:
$T_n = \frac{n(n + 1)}{2n} = \frac{n + 1}{2}$

(A) We know that the square of the sum of the first $n$ natural numbers is given by:
${\left\{ \frac{n(n + 1)}{2} \right\}^2} = {(1 + 2 + \dots + n)^2} = \sum_{r=1}^n {r^2} + 2\sum_{1 \le s < t \le n} {st}$
Therefore,the sum of products taken two at a time is:
$\sum_{1 \le s < t \le n} {st} = \frac{1}{2} \left[ {\left\{ \frac{n(n + 1)}{2} \right\}^2} - \sum_{r=1}^n {r^2} \right]$
Substituting the formula for the sum of squares $\sum_{r=1}^n {r^2} = \frac{n(n + 1)(2n + 1)}{6}$:
$= \frac{1}{2} \left[ \frac{n^2(n + 1)^2}{4} - \frac{n(n + 1)(2n + 1)}{6} \right]$
$= \frac{n(n + 1)}{2} \left[ \frac{n(n + 1)}{4} - \frac{2n + 1}{6} \right]$
$= \frac{n(n + 1)}{2} \left[ \frac{3n^2 + 3n - 4n - 2}{12} \right]$
$= \frac{n(n + 1)(3n^2 - n - 2)}{24}$
$= \frac{n(n + 1)(n - 1)(3n + 2)}{24}$

(A) The $n^{th}$ term of the series is given by $T_n = n(2n + 1)^2$.
Expanding this,we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.
To find the sum of $20$ terms,we calculate $S_{20} = \sum_{n=1}^{20} (4n^3 + 4n^2 + n) = 4\sum_{n=1}^{20} n^3 + 4\sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} n$.
Using the standard summation formulas:
$\sum_{n=1}^{N} n^3 = [\frac{N(N+1)}{2}]^2 = [\frac{20 \cdot 21}{2}]^2 = 210^2 = 44100$.
$\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6} = \frac{20 \cdot 21 \cdot 41}{6} = 2870$.
$\sum_{n=1}^{N} n = \frac{N(N+1)}{2} = \frac{20 \cdot 21}{2} = 210$.
Substituting these values: $S_{20} = 4(44100) + 4(2870) + 210 = 176400 + 11480 + 210 = 188090$.

(A) The sum of the first $n$ cubes is given by $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2$.
The sum of the first $n$ squares is given by $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.
Therefore,the ratio is:
$\frac{\sum_{k=1}^{12} k^3}{\sum_{k=1}^{12} k^2} = \frac{\left[ \frac{n(n+1)}{2} \right]^2}{\frac{n(n+1)(2n+1)}{6}} = \frac{n^2(n+1)^2}{4} \times \frac{6}{n(n+1)(2n+1)} = \frac{3n(n+1)}{2(2n+1)}$.
Substituting $n = 12$:
$\frac{3 \times 12 \times (12+1)}{2 \times (2 \times 12 + 1)} = \frac{3 \times 12 \times 13}{2 \times 25} = \frac{3 \times 6 \times 13}{25} = \frac{234}{25}$.

(A) The $n$-th term of the series is given by the product of three arithmetic progressions: $T_n = n(2n + 1)(3n + 2)$.
Expanding this,we get $T_n = n(6n^2 + 4n + 3n + 2) = 6n^3 + 7n^2 + 2n$.
The sum $S_n$ is $\sum_{k=1}^{n} T_k = 6\sum k^3 + 7\sum k^2 + 2\sum k$.
Using standard summation formulas:
$S_n = 6 \left[ \frac{n(n+1)}{2} \right]^2 + 7 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 2 \left[ \frac{n(n+1)}{2} \right]$.
Factoring out $\frac{n(n+1)}{6}$,we get:
$S_n = \frac{n(n+1)}{6} [9n(n+1) + 7(2n+1) + 6]$.
$S_n = \frac{n(n+1)}{6} [9n^2 + 9n + 14n + 7 + 6]$.
$S_n = \frac{n(n+1)(9n^2 + 23n + 13)}{6}$.

(D) The $n^{th}$ term of the series is given by $T_n = \frac{3^n - 1}{3^n} = 1 - \left(\frac{1}{3}\right)^n$.
The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n \left(1 - \left(\frac{1}{3}\right)^k\right)$.
$S_n = \sum_{k=1}^n 1 - \sum_{k=1}^n \left(\frac{1}{3}\right)^k$.
$S_n = n - \left[ \frac{\frac{1}{3}(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}} \right]$.
$S_n = n - \frac{\frac{1}{3}(1 - 3^{-n})}{\frac{2}{3}} = n - \frac{1}{2}(1 - 3^{-n})$.
$S_n = n - \frac{1}{2} + \frac{1}{2}3^{-n} = n + \frac{1}{2}(3^{-n} - 1)$.

(C) The sum of the squares of the first $n$ natural numbers is given by the standard formula:
$\sum\limits_{m = 1}^n {{m^2}} = 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{{n(n + 1)(2n + 1)}}{6}$
Therefore,the correct option is $C$.

(B) The $n^{th}$ term of the series is given by $T_n = n(n + 1) = n^2 + n$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k)$.
Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$,we get:
$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$
$S_n = \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 \right]$
$S_n = \frac{n(n+1)}{2} \left[ \frac{2n+1+3}{3} \right] = \frac{n(n+1)}{2} \left[ \frac{2n+4}{3} \right]$
$S_n = \frac{n(n+1) \times 2(n+2)}{2 \times 3} = \frac{1}{3}n(n + 1)(n + 2)$.

(B) The sum is given by $\sum_{n=11}^{20} n^3 = \sum_{n=1}^{20} n^3 - \sum_{n=1}^{10} n^3$.
Using the formula $\sum_{n=1}^{k} n^3 = \left[ \frac{k(k+1)}{2} \right]^2$:
Sum $= \left[ \frac{20(21)}{2} \right]^2 - \left[ \frac{10(11)}{2} \right]^2$
$= (210)^2 - (55)^2$
$= 44100 - 3025 = 41075$.
Since the last digit is $5$,the number $41075$ is divisible by $5$.
Since the last digit is $5$,it is an odd integer.
Therefore,the sum is an odd integer divisible by $5$.

(A) The $n^{th}$ term of the series is given by $T_n = n(2n + 1)^2$.
Expanding this,we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.
To find the sum $S_n$,we calculate $\sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (4k^3 + 4k^2 + k)$.
Using the standard summation formulas $\sum k^3 = \frac{n^2(n+1)^2}{4}$,$\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum k = \frac{n(n+1)}{2}$:
$S_n = 4 \cdot \frac{n^2(n+1)^2}{4} + 4 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$.
$S_n = n^2(n+1)^2 + \frac{2n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$.
Factoring out $\frac{n(n+1)}{6}$,we get:
$S_n = \frac{n(n+1)}{6} [6n(n+1) + 4(2n+1) + 3]$.
$S_n = \frac{n(n+1)}{6} [6n^2 + 6n + 8n + 4 + 3]$.
$S_n = \frac{n(n+1)}{6} (6n^2 + 14n + 7)$.

(B) Given the $n^{th}$ term $T_n = 3 + n(n - 1) = n^2 - n + 3$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 - k + 3)$.
Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum k = \frac{n(n+1)}{2}$,and $\sum 1 = n$:
$S_n = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} + 3n$.
$S_n = \frac{n(n+1)(2n+1) - 3n(n+1) + 18n}{6}$.
$S_n = \frac{n[(n+1)(2n+1) - 3(n+1) + 18]}{6}$.
$S_n = \frac{n[2n^2 + 3n + 1 - 3n - 3 + 18]}{6}$.
$S_n = \frac{n[2n^2 + 16]}{6} = \frac{2n(n^2 + 8)}{6} = \frac{n(n^2 + 8)}{3} = \frac{n^3 + 8n}{3}$.

(D) The $k$-th term of the series is $T_k = k(2n - (2k - 1)) = k(2n - 2k + 1) = 2nk - 2k^2 + k$.
Summing from $k = 1$ to $n$:
$S = \sum_{k=1}^{n} (2nk - 2k^2 + k) = 2n \sum_{k=1}^{n} k - 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$.
Using standard summation formulas:
$S = 2n \frac{n(n+1)}{2} - 2 \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$.
$S = n^2(n+1) - \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$.
Factor out $\frac{n(n+1)}{6}$:
$S = \frac{n(n+1)}{6} [6n - 2(2n+1) + 3] = \frac{n(n+1)}{6} [6n - 4n - 2 + 3] = \frac{n(n+1)(2n+1)}{6}$.