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

(A) The $n^{th}$ group contains $n$ terms in an arithmetic progression with a common difference of $1$.
The first term of the $n^{th}$ group,$T_n$,is the sum of the number of terms in the previous $(n-1)$ groups plus $1$.
The number of terms in the first $(n-1)$ groups is $1 + 2 + 3 + \dots + (n-1) = \frac{(n-1)n}{2}$.
Thus,the first term of the $n^{th}$ group is $T_n = \frac{n(n-1)}{2} + 1 = \frac{n^2 - n + 2}{2}$.
The sum of the $n$ terms in the $n^{th}$ group is given by the formula for the sum of an arithmetic progression: $S_n = \frac{n}{2} [2a + (n-1)d]$,where $a = T_n$ and $d = 1$.
$S_n = \frac{n}{2} [2(\frac{n^2 - n + 2}{2}) + (n-1)(1)]$
$S_n = \frac{n}{2} [n^2 - n + 2 + n - 1]$
$S_n = \frac{n}{2} [n^2 + 1] = \frac{1}{2} [n(n^2 + 1)]$.

(B) Let the series be $S_n = 1 + 6 + 13 + 22 + 33 + \dots + T_n$.
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 $n=1, T_1 = a + b + c = 1$.
For $n=2, T_2 = 4a + 2b + c = 6$.
For $n=3, T_3 = 9a + 3b + c = 13$.
Subtracting the equations:
$(4a + 2b + c) - (a + b + c) = 6 - 1 \implies 3a + b = 5$.
$(9a + 3b + c) - (4a + 2b + c) = 13 - 6 \implies 5a + b = 7$.
Subtracting these results: $(5a + b) - (3a + b) = 7 - 5 \implies 2a = 2 \implies a = 1$.
Substituting $a=1$ into $3a + b = 5$,we get $3(1) + b = 5 \implies b = 2$.
Substituting $a=1, b=2$ into $a + b + c = 1$,we get $1 + 2 + c = 1 \implies c = -2$.
Thus,$T_n = n^2 + 2n - 2$.
Now,$S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^2 + 2k - 2) = \sum k^2 + 2\sum k - \sum 2$.
$S_n = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} - 2n$.
$S_n = \frac{n(2n^2 + 3n + 1) + 6n(n+1) - 12n}{6} = \frac{n(2n^2 + 3n + 1 + 6n + 6 - 12)}{6} = \frac{n(2n^2 + 9n - 5)}{6}$.

(A) The given series is $S_n = 1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots + T_n$.
When $n$ is even,the sum is given as $S_n = \frac{n(n+1)^2}{2}$.
When $n$ is odd,the $(n-1)$-th term is even.
Thus,$S_n = S_{n-1} + T_n$.
Since $n-1$ is even,$S_{n-1} = \frac{(n-1)((n-1)+1)^2}{2} = \frac{(n-1)n^2}{2}$.
The $n$-th term $T_n$ for odd $n$ is $n^2$.
Therefore,$S_n = \frac{(n-1)n^2}{2} + n^2 = \frac{n^2(n-1+2)}{2} = \frac{n^2(n+1)}{2}$.

(D) The $n$-th term of the sequence is given by $T_n = (2n - 1)(2n + 1)(2n + 3)$.
Expanding this,we get $T_n = (4n^2 - 1)(2n + 3) = 8n^3 + 12n^2 - 2n - 3$.
The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} T_k = 8\sum k^3 + 12\sum k^2 - 2\sum k - \sum 3$.
Using the standard summation formulas:
$S_n = 8 \left[ \frac{n(n+1)}{2} \right]^2 + 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 2 \left[ \frac{n(n+1)}{2} \right] - 3n$.
$S_n = 2n^2(n^2 + 2n + 1) + 2n(2n^2 + 3n + 1) - n(n+1) - 3n$.
$S_n = 2n^4 + 4n^3 + 2n^2 + 4n^3 + 6n^2 + 2n - n^2 - n - 3n$.
$S_n = 2n^4 + 8n^3 + 7n^2 - 2n$.
The arithmetic mean is $\frac{S_n}{n} = \frac{2n^4 + 8n^3 + 7n^2 - 2n}{n} = 2n^3 + 8n^2 + 7n - 2$.

(A) Let the sum be $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ $(1)$
Multiply by $\frac{1}{3}$: $\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \frac{10}{3^4} + \dots$ $(2)$
Subtracting $(2)$ from $(1)$:
$S - \frac{1}{3}S = 1 + (\frac{2}{3} - \frac{1}{3}) + (\frac{6}{3^2} - \frac{2}{3^2}) + (\frac{10}{3^3} - \frac{6}{3^3}) + (\frac{14}{3^4} - \frac{10}{3^4}) + \dots$
$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{3^2} + \frac{4}{3^3} + \frac{4}{3^4} + \dots$
$\frac{2}{3}S = 1 + \frac{1}{3} + [\frac{4}{3^2} + \frac{4}{3^3} + \frac{4}{3^4} + \dots]$
The term in the bracket is an infinite geometric series with $a = \frac{4}{9}$ and $r = \frac{1}{3}$.
Sum $= \frac{a}{1-r} = \frac{4/9}{1 - 1/3} = \frac{4/9}{2/3} = \frac{4}{9} \times \frac{3}{2} = \frac{2}{3}$.
So,$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{2}{3} = 1 + 1 = 2$.
$S = 2 \times \frac{3}{2} = 3$.

(C) Let $S_{20} = 0.7 + 0.77 + 0.777 + \dots$ up to $20$ terms.
$S_{20} = 7[0.1 + 0.11 + 0.111 + \dots \text{ up to } 20 \text{ terms}]$.
Multiply and divide by $9$:
$S_{20} = \frac{7}{9}[0.9 + 0.99 + 0.999 + \dots \text{ up to } 20 \text{ terms}]$.
$S_{20} = \frac{7}{9}[(1 - 10^{-1}) + (1 - 10^{-2}) + (1 - 10^{-3}) + \dots + (1 - 10^{-20})]$.
$S_{20} = \frac{7}{9}[20 - (10^{-1} + 10^{-2} + \dots + 10^{-20})]$.
The sum of the geometric series inside the bracket is $\frac{0.1(1 - 0.1^{20})}{1 - 0.1} = \frac{0.1(1 - 10^{-20})}{0.9} = \frac{1}{9}(1 - 10^{-20})$.
$S_{20} = \frac{7}{9}[20 - \frac{1}{9}(1 - 10^{-20})] = \frac{7}{9}[\frac{180 - 1 + 10^{-20}}{9}] = \frac{7}{81}(179 + 10^{-20})$.

(C) The $n^{th}$ term of the series is given by $T_n = \frac{\sum_{k=1}^n k^3}{\sum_{k=1}^n (2k-1)}$.
We know that $\sum_{k=1}^n k^3 = \left[\frac{n(n+1)}{2}\right]^2$ and the sum of the first $n$ odd numbers is $\sum_{k=1}^n (2k-1) = n^2$.
Thus,$T_n = \frac{\left[\frac{n(n+1)}{2}\right]^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4}$.
We need to find the sum of the first $9$ terms,$S_9 = \sum_{n=1}^9 T_n = \sum_{n=1}^9 \frac{(n+1)^2}{4}$.
$S_9 = \frac{1}{4} \sum_{n=1}^9 (n+1)^2 = \frac{1}{4} (2^2 + 3^2 + \dots + 10^2)$.
Adding and subtracting $1^2$,we get $S_9 = \frac{1}{4} \left[ \sum_{k=1}^{10} k^2 - 1^2 \right]$.
Using the formula $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$,for $m=10$:
$S_9 = \frac{1}{4} \left[ \frac{10(11)(21)}{6} - 1 \right] = \frac{1}{4} [385 - 1] = \frac{384}{4} = 96$.

(D) The series is $S = \left(\frac{8}{5}\right)^2 + \left(\frac{12}{5}\right)^2 + \left(\frac{16}{5}\right)^2 + \left(\frac{20}{5}\right)^2 + \dots$ up to $10$ terms.
This can be written as $S = \frac{1}{25} \left( 8^2 + 12^2 + 16^2 + 20^2 + \dots + (4(n+1))^2 \right)$ for $n=1$ to $10$.
$S = \frac{16}{25} \sum_{n=1}^{10} (n+1)^2$.
Let $k = n+1$,then the sum is $\frac{16}{25} \sum_{k=2}^{11} k^2$.
Using the formula $\sum_{k=1}^{N} k^2 = \frac{N(N+1)(2N+1)}{6}$,we have $\sum_{k=1}^{11} k^2 = \frac{11(12)(23)}{6} = 11 \times 2 \times 23 = 506$.
Since the sum starts from $k=2$,we subtract $1^2 = 1$: $506 - 1 = 505$.
So,$S = \frac{16}{25} \times 505 = \frac{16 \times 101}{5} = \frac{16}{5} \times 101$.
Given $S = \frac{16}{5}m$,we get $m = 101$.

(A) The series is $a_n = n^2$ if $n$ is odd,and $a_n = 2n^2$ if $n$ is even.
$B - 2A = \sum_{n=1}^{40} a_n - 2\sum_{n=1}^{20} a_n = \sum_{n=21}^{40} a_n - \sum_{n=1}^{20} a_n$.
For $n \in [21, 40]$,let $k = n-20$,so $n = k+20$. The terms are $(k+20)^2$ if $k$ is odd,and $2(k+20)^2$ if $k$ is even.
$B - 2A = \sum_{k=1}^{20} (a_{k+20} - a_k)$.
If $k$ is odd,$a_{k+20} - a_k = (k+20)^2 - k^2 = 40k + 400$.
If $k$ is even,$a_{k+20} - a_k = 2(k+20)^2 - 2k^2 = 2(40k + 400) = 80k + 800$.
Summing these for $k=1$ to $20$ ($10$ odd and $10$ even values of $k$):
$B - 2A = \sum_{k \in \text{odd}} (40k + 400) + \sum_{k \in \text{even}} (80k + 800)$.
$= 40 \sum_{k \in \text{odd}} k + 4000 + 80 \sum_{k \in \text{even}} k + 8000$.
$= 40(1+3+\dots+19) + 80(2+4+\dots+20) + 12000$.
$= 40(100) + 80(110) + 12000 = 4000 + 8800 + 12000 = 24800$.
Since $B - 2A = 100\lambda$,$100\lambda = 24800$,so $\lambda = 248$.

(C) The given series is $S = 2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots \infty$.
We can rearrange the terms as:
$S = (1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \infty) + (1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots \infty)$.
Both parts are infinite geometric series with the sum formula $S_{\infty} = \frac{a}{1-r}$.
For the first series,$a = 1$ and $r = 1/2$,so $S_1 = \frac{1}{1 - 1/2} = 2$.
For the second series,$a = 1$ and $r = 1/3$,so $S_2 = \frac{1}{1 - 1/3} = \frac{1}{2/3} = 3/2$.
Thus,the total sum is $S = S_1 + S_2 = 2 + 3/2 = 7/2$.

(D) The number of terms in the $k^{th}$ row is $2k$.
The total number of terms in the first $(n-1)$ rows is $2 + 4 + 6 + \dots + 2(n-1) = 2 \times \frac{(n-1)n}{2} = n(n-1) = n^2 - n$.
The first term of the $n^{th}$ row is the $(n^2 - n + 1)^{th}$ odd number.
The $m^{th}$ odd number is $2m - 1$.
So,the first term $a$ of the $n^{th}$ row is $2(n^2 - n + 1) - 1 = 2n^2 - 2n + 2 - 1 = 2n^2 - 2n + 1$.
The $n^{th}$ row is an arithmetic progression with $2n$ terms,first term $a = 2n^2 - 2n + 1$,and common difference $d = 2$.
The sum $S_n$ of the $n^{th}$ row is given by $S_n = \frac{2n}{2} [2a + (2n - 1)d]$.
$S_n = n [2(2n^2 - 2n + 1) + (2n - 1)2] = n [4n^2 - 4n + 2 + 4n - 2] = n [4n^2] = 4n^3$.

(A) Let $T_k$ be the $k$-th term of the series,then $T_k = 3k(k + 1) = 3k^2 + 3k$.
If $S_n$ denotes the sum of the first $n$ terms,then $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (3k^2 + 3k)$.
$S_n = 3 \sum_{k=1}^n k^2 + 3 \sum_{k=1}^n k$.
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:
$S_n = 3 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 3 \left[ \frac{n(n+1)}{2} \right]$.
$S_n = \frac{n(n+1)(2n+1)}{2} + \frac{3n(n+1)}{2} = \frac{n(n+1)}{2} [ (2n+1) + 3 ]$.
$S_n = \frac{n(n+1)}{2} [ 2n + 4 ] = \frac{n(n+1)}{2} \cdot 2(n + 2) = n(n+1)(n+2)$.

(A) Let $S = 0.1 + 0.01 + 0.001 + \dots$
This is an infinite geometric series with first term $a = 0.1$ and common ratio $r = 0.1$.
The sum $S = \frac{a}{1-r} = \frac{0.1}{1-0.1} = \frac{0.1}{0.9} = \frac{1}{9}$.
Now,the expression is $(0.05)^{\log_{\sqrt{20}}(1/9)}$.
Note that $0.05 = \frac{5}{100} = \frac{1}{20}$.
Also,$\sqrt{20} = 20^{1/2}$.
Using the property $\log_{a^n}(b) = \frac{1}{n} \log_a(b)$,we have $\log_{20^{1/2}}(1/9) = \frac{1}{1/2} \log_{20}(1/9) = 2 \log_{20}(1/9) = \log_{20}(1/9)^2 = \log_{20}(1/81)$.
So,the expression becomes $(1/20)^{\log_{20}(1/81)}$.
Using the property $a^{\log_a(x)} = x$,we rewrite $(1/20)^{\log_{20}(1/81)}$ as $(20^{-1})^{\log_{20}(1/81)} = (20^{\log_{20}(1/81)})^{-1} = (1/81)^{-1} = 81$.

(A) The given expression is the Binet's formula for the $n$-th Fibonacci number,where $u_n = F_n$.
Let $\alpha = \frac{1 + \sqrt{5}}{2}$ and $\beta = \frac{1 - \sqrt{5}}{2}$.
Then $u_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}$.
We know that $\alpha$ and $\beta$ are roots of the quadratic equation $x^2 - x - 1 = 0$,so $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$.
For $n \ge 1$,$u_{n+1} = \frac{\alpha^{n+1} - \beta^{n+1}}{\sqrt{5}}$.
$u_n + u_{n-1} = \frac{1}{\sqrt{5}} [(\alpha^n - \beta^n) + (\alpha^{n-1} - \beta^{n-1})] = \frac{1}{\sqrt{5}} [\alpha^{n-1}(\alpha + 1) - \beta^{n-1}(\beta + 1)]$.
Since $\alpha + 1 = \alpha^2$ and $\beta + 1 = \beta^2$,we get $u_n + u_{n-1} = \frac{\alpha^{n+1} - \beta^{n+1}}{\sqrt{5}} = u_{n+1}$.
Thus,the recurrence relation is $u_{n+1} = u_n + u_{n-1}$.

(B) We are given the equation $\sum_{r=1}^{n}r^3 - \sum_{p=1}^{n}\sum_{m=1}^{p}\sum_{r=1}^{m}1 = 80$.
First,we know that $\sum_{r=1}^{n}r^3 = \left[ \frac{n(n+1)}{2} \right]^2$.
Next,consider the triple summation: $\sum_{p=1}^{n}\sum_{m=1}^{p}\sum_{r=1}^{m}1 = \sum_{p=1}^{n}\sum_{m=1}^{p} m = \sum_{p=1}^{n} \frac{p(p+1)}{2} = \frac{1}{2} \left[ \sum_{p=1}^{n} p^2 + \sum_{p=1}^{n} p \right]$.
Using standard summation formulas: $\frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right] = \frac{n(n+1)}{4} \left[ \frac{2n+1}{3} + 1 \right] = \frac{n(n+1)}{4} \left[ \frac{2n+4}{3} \right] = \frac{n(n+1)(n+2)}{6} = \binom{n+2}{3}$.
Substituting back into the original equation: $\left[ \frac{n(n+1)}{2} \right]^2 - \frac{n(n+1)(n+2)}{6} = 80$.
For $n=4$: $\left[ \frac{4(5)}{2} \right]^2 - \frac{4(5)(6)}{6} = 10^2 - 20 = 100 - 20 = 80$.
Thus,$n=4$ is the correct value.

(B) We are given the triple summation: $\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j 1 } } = 560$.
First,evaluate the innermost sum: $\sum\limits_{k = 1}^j 1 = j$.
Now,the expression becomes: $\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i j } = 560$.
Next,evaluate the middle sum: $\sum\limits_{j = 1}^i j = \frac{i(i+1)}{2}$.
Now,the expression becomes: $\sum\limits_{i = 1}^n \frac{i(i+1)}{2} = 560$.
This can be written as: $\frac{1}{2} \left[ \sum\limits_{i=1}^n i^2 + \sum\limits_{i=1}^n i \right] = 560$.
Using the standard summation formulas: $\frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right] = 560$.
Factoring out $\frac{n(n+1)}{2}$: $\frac{1}{2} \cdot \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 \right] = 560$.
$\frac{n(n+1)}{4} \left[ \frac{2n+4}{3} \right] = 560$.
$\frac{n(n+1) \cdot 2(n+2)}{12} = 560$.
$\frac{n(n+1)(n+2)}{6} = 560$.
$n(n+1)(n+2) = 560 \times 6 = 3360$.
We look for three consecutive integers whose product is $3360$. Since $14 \times 15 \times 16 = 3360$,we have $n = 14$.

(B) Let $S = \sum\limits_{n = 1}^\infty \sum\limits_{k = 1}^{n - 1} \frac{k}{2^{n+k}} = \sum\limits_{n = 1}^\infty \frac{1}{2^n} \sum\limits_{k = 1}^{n - 1} \frac{k}{2^k}$.
We know that $\sum\limits_{k = 1}^{n - 1} k x^k = \frac{x(1 - nx^{n-1} + (n-1)x^n)}{(1-x)^2}$.
For $x = \frac{1}{2}$,$\sum\limits_{k = 1}^{n - 1} \frac{k}{2^k} = \frac{1/2(1 - n(1/2)^{n-1} + (n-1)(1/2)^n)}{(1/2)^2} = 2(1 - \frac{n}{2^{n-1}} + \frac{n-1}{2^n}) = 2 - \frac{2n}{2^{n-1}} + \frac{n-1}{2^{n-1}} = 2 - \frac{n+1}{2^{n-1}}$.
Substituting this back,$S = \sum\limits_{n = 1}^\infty \frac{1}{2^n} (2 - \frac{n+1}{2^{n-1}}) = \sum\limits_{n = 1}^\infty (\frac{1}{2^{n-1}} - \frac{n+1}{2^{2n-1}})$.
$S = \sum\limits_{n = 1}^\infty \frac{1}{2^{n-1}} - \sum\limits_{n = 1}^\infty \frac{n+1}{2^{2n-1}} = 2 - 2 \sum\limits_{n = 1}^\infty \frac{n+1}{4^n}$.
Using $\sum\limits_{n=1}^\infty nx^n = \frac{x}{(1-x)^2}$ and $\sum\limits_{n=1}^\infty x^n = \frac{x}{1-x}$,we have $\sum\limits_{n=1}^\infty (n+1)x^n = \frac{x}{(1-x)^2} + \frac{x}{1-x} = \frac{x + x(1-x)}{(1-x)^2} = \frac{2x-x^2}{(1-x)^2}$.
For $x = 1/4$,$\sum\limits_{n=1}^\infty (n+1)(1/4)^n = \frac{2(1/4) - (1/16)}{(3/4)^2} = \frac{1/2 - 1/16}{9/16} = \frac{7/16}{9/16} = \frac{7}{9}$.
Thus,$S = 2 - 2(\frac{7}{9}) = 2 - \frac{14}{9} = \frac{4}{9}$.

(A) Let $S = \sum_{n=0}^{\infty} \frac{(n+1)^2}{7^n} = \frac{1^2}{7^0} + \frac{2^2}{7^1} + \frac{3^2}{7^2} + \frac{4^2}{7^3} + \dots$
Multiply by $\frac{1}{7}$:
$\frac{S}{7} = \frac{1^2}{7^1} + \frac{2^2}{7^2} + \frac{3^2}{7^3} + \dots$
Subtract the two equations:
$S - \frac{S}{7} = 1 + \frac{2^2 - 1^2}{7^1} + \frac{3^2 - 2^2}{7^2} + \frac{4^2 - 3^2}{7^3} + \dots$
$\frac{6S}{7} = 1 + \frac{3}{7} + \frac{5}{7^2} + \frac{7}{7^3} + \dots$
This is an arithmetico-geometric series. Let $T = 1 + \frac{3}{7} + \frac{5}{7^2} + \frac{7}{7^3} + \dots$
$\frac{T}{7} = \frac{1}{7} + \frac{3}{7^2} + \frac{5}{7^3} + \dots$
Subtracting these:
$T - \frac{T}{7} = 1 + \frac{2}{7} + \frac{2}{7^2} + \frac{2}{7^3} + \dots$
$\frac{6T}{7} = 1 + \frac{2/7}{1 - 1/7} = 1 + \frac{2/7}{6/7} = 1 + \frac{1}{3} = \frac{4}{3}$
$T = \frac{4}{3} \times \frac{7}{6} = \frac{14}{9}$
Since $\frac{6S}{7} = T$,we have $\frac{6S}{7} = \frac{14}{9} \Rightarrow S = \frac{14}{9} \times \frac{7}{6} = \frac{98}{54} = \frac{49}{27}$.

(D) Let $n = 2015$. The general term of the series is $T_k = k(n - k + 1)$ for $k = 1, 2, \dots, n$.
We need to find the sum $S = \sum_{k=1}^{n} k(n - k + 1) = \sum_{k=1}^{n} (k(n+1) - k^2)$.
Using the summation formulas $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$S = (n+1) \sum_{k=1}^{n} k - \sum_{k=1}^{n} k^2 = (n+1) \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$.
Factor out $\frac{n(n+1)}{6}$:
$S = \frac{n(n+1)}{6} [3(n+1) - (2n+1)] = \frac{n(n+1)}{6} [3n + 3 - 2n - 1] = \frac{n(n+1)(n+2)}{6}$.
Substituting $n = 2015$:
$S = \frac{2015 \times 2016 \times 2017}{6} = 2015 \times 336 \times 2017$.
Comparing this with the options,none of the given options match the result $2015 \times 336 \times 2017$ exactly as written in the choices provided.

(A) The given series is $S = \frac{1}{9} + \frac{1}{18} + \frac{1}{30} + \frac{1}{45} + \frac{1}{63} + \dots \infty$.
Multiplying by $2$,we get $2S = \frac{2}{9} + \frac{2}{18} + \frac{2}{30} + \frac{2}{45} + \frac{2}{63} + \dots \infty$.
This can be written as $2S = \frac{2}{3 \times 3} + \frac{2}{3 \times 6} + \frac{2}{6 \times 5} + \dots$ which is not immediately helpful.
Let us rewrite the terms as $T_n = \frac{1}{\frac{3n(n+2)}{2}} = \frac{2}{3n(n+2)}$.
Using partial fractions,$T_n = \frac{2}{3} \times \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+2} \right) = \frac{1}{3} \left( \frac{1}{n} - \frac{1}{n+2} \right)$.
For $n=1, 2, 3, \dots$,the terms are $\frac{1}{3} (1 - \frac{1}{3}), \frac{1}{3} (\frac{1}{2} - \frac{1}{4}), \frac{1}{3} (\frac{1}{3} - \frac{1}{5}), \dots$.
Sum $S = \frac{1}{3} [ (1 - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{4} - \frac{1}{6}) + \dots ]$.
Most terms cancel out,leaving $S = \frac{1}{3} (1 + \frac{1}{2}) = \frac{1}{3} (\frac{3}{2}) = \frac{1}{2}$.
Wait,re-evaluating the series: $9, 18, 30, 45, 63$. The differences are $9, 12, 15, 18$.
These are $3 \times 3, 3 \times 6, 3 \times 10, 3 \times 15, 3 \times 21$.
The denominators are $3 \times \frac{n(n+1)}{2}$ where $n=2, 3, 4, 5, 6$.
$T_n = \frac{2}{3n(n+1)}$.
$S = \sum_{n=2}^{\infty} \frac{2}{3} (\frac{1}{n} - \frac{1}{n+1}) = \frac{2}{3} [(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots] = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$.
Thus,the correct option is $A$.

(D) The number of terms in successive groups are $1, 2, 3, \dots, n$. Thus,the $n^{th}$ group contains $n$ terms in an arithmetic progression with a common difference $d = 2$.
First,we find the first term of the $n^{th}$ group. Let $a_n$ be the first term of the $n^{th}$ group. The sequence of first terms is $1, 3, 7, 13, \dots$. The differences between consecutive terms are $2, 4, 6, \dots$,which form an arithmetic progression.
Let $a_n = a_1 + \sum_{k=1}^{n-1} (2k) = 1 + 2 \cdot \frac{(n-1)n}{2} = 1 + n^2 - n = n^2 - n + 1$.
The $n^{th}$ group is an arithmetic progression with $n$ terms,first term $a = n^2 - n + 1$,and common difference $d = 2$.
The sum of the $n^{th}$ group is $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $S_n = \frac{n}{2} [2(n^2 - n + 1) + (n - 1)2] = \frac{n}{2} [2n^2 - 2n + 2 + 2n - 2] = \frac{n}{2} [2n^2] = n^3$.

(A) The given quadratic equation is $x^2 - r^2(r + 1)x + r^5 = 0$.
By Vieta's formulas,the sum of roots $\alpha_r + \beta_r = r^2(r + 1) = r^3 + r^2$ and the product of roots $\alpha_r \beta_r = r^5$.
Since $\alpha_r \beta_r = r^2 \cdot r^3 = r^5$ and $\alpha_r + \beta_r = r^2 + r^3$,the roots are $\alpha_r = r^2$ and $\beta_r = r^3$ (given $\alpha_r < \beta_r$ for $r > 1$).
We need to calculate $S = \sum_{r=1}^{n} (3\alpha_r + 2\beta_r) = \sum_{r=1}^{n} (3r^2 + 2r^3)$.
Using the standard summation formulas $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4}$:
$S = 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n^2(n+1)^2}{4} = \frac{n(n+1)(2n+1)}{2} + \frac{n^2(n+1)^2}{2}$.
Factoring out $\frac{n(n+1)}{2}$:
$S = \frac{n(n+1)}{2} [ (2n+1) + n(n+1) ] = \frac{n(n+1)}{2} [ 2n + 1 + n^2 + n ] = \frac{n(n+1)(n^2 + 3n + 1)}{2}$.

(B) Given $x_n = \frac{2n^2 + n + 1}{2n^2 - 3n + 2}$. Let $f(n) = 2n^2 - 3n + 2$. Then $f(n-1) = 2(n-1)^2 - 3(n-1) + 2 = 2(n^2 - 2n + 1) - 3n + 3 + 2 = 2n^2 - 7n + 7$. This does not match the numerator directly. Let us re-evaluate $x_n = \frac{2n^2 + n + 1}{2(n-1)^2 - 3(n-1) + 2} = \frac{2n^2 + n + 1}{2n^2 - 7n + 7}$.
Actually,observing the product $\prod_{i=1}^r x_i$: for $r=1$,$x_1 = \frac{2+1+1}{2-3+2} = \frac{4}{1} = 4$. For $r=2$,$x_1 x_2 = 4 \times \frac{8+2+1}{8-6+2} = 4 \times \frac{11}{4} = 11$. For $r=3$,$x_1 x_2 x_3 = 11 \times \frac{18+3+1}{18-9+2} = 11 \times \frac{22}{11} = 22$. The product is $2r^2 + r + 1$.
The second term is $2 \sum_{i=1}^r (2i - 1) = 2(r^2) = 2r^2$.
Thus,the expression inside the summation is $(2r^2 + r + 1) - 2r^2 = r + 1$.
Finally,$\sum_{r=1}^n (r + 1) = \sum_{r=1}^n r + \sum_{r=1}^n 1 = \frac{n(n+1)}{2} + n = \frac{n^2 + n + 2n}{2} = \frac{n(n+3)}{2}$.

(A) To climb an $n$-step staircase,the person can reach the $n$-th step either from the $(n-1)$-th step (by taking one step) or from the $(n-2)$-th step (by taking two steps).
Therefore,the recurrence relation is $C_n = C_{n-1} + C_{n-2}$ for $n \ge 2$.
This is the definition of the Fibonacci sequence.
Given the recurrence relation $C_n = C_{n-1} + C_{n-2}$,we can substitute $n = 20$ to get $C_{20} = C_{19} + C_{18}$.
Thus,$C_{18} + C_{19} = C_{20}$.

(C) Let the common ratio be $r$. Since $b_1 + b_2 = 1$,we have $b_1(1 + r) = 1$,so $b_1 = \frac{1}{1 + r}$.
The sum of the infinite geometric series is given by $\frac{b_1}{1 - r} = 2$.
Substituting $b_1 = \frac{1}{1 + r}$,we get $\frac{1}{(1 + r)(1 - r)} = 2$,which simplifies to $\frac{1}{1 - r^2} = 2$.
This implies $1 - r^2 = \frac{1}{2}$,so $r^2 = \frac{1}{2}$,which means $r = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$.
Given $b_2 = b_1 r < 0$,and since $b_1 = \frac{1}{1 + r}$,we have $b_2 = \frac{r}{1 + r} < 0$. This condition is satisfied when $r$ is negative,so $r = -\frac{\sqrt{2}}{2}$.
Thus,$b_1 = \frac{1}{1 - \frac{\sqrt{2}}{2}} = \frac{1}{\frac{2 - \sqrt{2}}{2}} = \frac{2}{2 - \sqrt{2}} = \frac{2(2 + \sqrt{2})}{4 - 2} = 2 + \sqrt{2}$.

(C) The given series is $\left(\frac{11}{7}\right)^2 + \left(\frac{12}{7}\right)^2 + \left(\frac{13}{7}\right)^2 + \dots + \left(\frac{21}{7}\right)^2$.
The sum of the first $11$ terms is $S = \sum_{n=11}^{21} \left(\frac{n}{7}\right)^2 = \frac{1}{49} \sum_{n=11}^{21} n^2$.
Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we have:
$S = \frac{1}{49} \left[ \sum_{n=1}^{21} n^2 - \sum_{n=1}^{10} n^2 \right]$
$S = \frac{1}{49} \left[ \frac{21 \times 22 \times 43}{6} - \frac{10 \times 11 \times 21}{6} \right]$
$S = \frac{1}{49} \times \frac{21}{6} \times [ (22 \times 43) - (10 \times 11) ]$
$S = \frac{1}{49} \times \frac{7}{2} \times [ 946 - 110 ]$
$S = \frac{1}{14} \times 836 = \frac{418}{7} = \frac{11}{7} \times 38$.
Comparing this with $\frac{11}{7}\lambda$,we get $\lambda = 38$.

(C) The $n^{th}$ term of a sequence is given by $T_n = S_n - S_{n-1}$ for $n > 1$.
Given $S_n = 4n^2 + 6n$.
Then $S_{n-1} = 4(n-1)^2 + 6(n-1) = 4(n^2 - 2n + 1) + 6n - 6 = 4n^2 - 8n + 4 + 6n - 6 = 4n^2 - 2n - 2$.
$T_n = (4n^2 + 6n) - (4n^2 - 2n - 2) = 8n + 2$.
For $n = 15$,$T_{15} = 8(15) + 2 = 120 + 2 = 122$.

(D) Given the recurrence relation: $a_{n+1} - a_n = n - 2$.
Summing this from $n = 1$ to $50$:
$\sum_{n=1}^{50} (a_{n+1} - a_n) = \sum_{n=1}^{50} (n - 2)$
$a_{51} - a_1 = \sum_{n=1}^{50} n - \sum_{n=1}^{50} 2$
$a_{51} - 5 = \frac{50 \times 51}{2} - (50 \times 2)$
$a_{51} - 5 = 1275 - 100$
$a_{51} - 5 = 1175$
$a_{51} = 1180$.

(A) The $r$-th term of the series is $t_r = 1^2 + 3^2 + 5^2 + \dots + (2r - 1)^2$.
$t_r = \sum_{k=1}^r (2k - 1)^2 = \sum_{k=1}^r (4k^2 - 4k + 1)$.
Using standard summation formulas:
$t_r = 4 \cdot \frac{r(r+1)(2r+1)}{6} - 4 \cdot \frac{r(r+1)}{2} + r$.
Simplifying this expression:
$t_r = \frac{2r(r+1)(2r+1)}{3} - 2r(r+1) + r = \frac{2r(2r^2 + 3r + 1) - 6r^2 - 6r + 3r}{3} = \frac{4r^3 + 6r^2 + 2r - 6r^2 - 3r}{3} = \frac{4r^3 - r}{3}$.
The sum of $n$ terms is $S_n = \sum_{r=1}^n t_r = \sum_{r=1}^n \frac{4r^3 - r}{3} = \frac{4}{3} \sum_{r=1}^n r^3 - \frac{1}{3} \sum_{r=1}^n r$.
$S_n = \frac{4}{3} \left[ \frac{n(n+1)}{2} \right]^2 - \frac{1}{3} \left[ \frac{n(n+1)}{2} \right]$.
$S_n = \frac{n(n+1)}{6} [2n(n+1) - 1] = \frac{1}{6} n(n+1)(2n^2 + 2n - 1)$.

(D) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = \sin \theta$.
Since the sum to infinity is $S = \frac{a}{1-r}$,we have $\frac{1}{1 - \sin \theta} = 4 + 2\sqrt{3}$.
Taking the reciprocal,$1 - \sin \theta = \frac{1}{4 + 2\sqrt{3}}$.
Rationalizing the denominator: $1 - \sin \theta = \frac{4 - 2\sqrt{3}}{(4 + 2\sqrt{3})(4 - 2\sqrt{3})} = \frac{4 - 2\sqrt{3}}{16 - 12} = \frac{4 - 2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$.
Thus,$\sin \theta = \frac{\sqrt{3}}{2}$.
Given $0 < \theta < \pi$ and $\theta \neq \frac{\pi}{2}$,the possible values for $\theta$ are $\frac{\pi}{3}$ and $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

(C) Let $S = 3 + \frac{3+d}{4} + \frac{3+2d}{4^2} + \dots \infty$ $(1)$
Multiply by $\frac{1}{4}$:
$\frac{S}{4} = \frac{3}{4} + \frac{3+d}{4^2} + \frac{3+2d}{4^3} + \dots \infty$ $(2)$
Subtract $(2)$ from $(1)$:
$S - \frac{S}{4} = 3 + \frac{d}{4} + \frac{d}{4^2} + \frac{d}{4^3} + \dots \infty$
$\frac{3S}{4} = 3 + \frac{d/4}{1 - 1/4} = 3 + \frac{d/4}{3/4} = 3 + \frac{d}{3}$
Given $S = 8$,so $\frac{3(8)}{4} = 3 + \frac{d}{3}$
$6 = 3 + \frac{d}{3}$ $\Rightarrow 3 = \frac{d}{3}$ $\Rightarrow d = 9$

(C) The given sum is $S = \sum_{n=1}^{2009} n(2010 - n)$.
This can be written as $S = 2010 \sum_{n=1}^{2009} n - \sum_{n=1}^{2009} n^2$.
We know $\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$,so $\sum_{n=1}^{2009} n = \frac{2009 \times 2010}{2} = 2009 \times 1005$.
Thus,$S = 2010(2009 \times 1005) - (2009)(335)(4019)$.
$S = (2009)(2010 \times 1005) - (2009)(335)(4019)$.
$S = (2009)(335) \times [3 \times 2010 - 4019]$.
$S = (2009)(335) \times [6030 - 4019]$.
$S = (2009)(335) \times 2011$.
Comparing this with $(2009)(335)(x)$,we get $x = 2011$.

(C) Let the first term be $b$ and the common ratio be $r$.
For an infinite $G.P.$,the sum $S$ is given by $S = \frac{b}{1 - r}$,where $|r| < 1$.
Given $S = 5$,we have $\frac{b}{1 - r} = 5$.
This implies $b = 5(1 - r)$.
Since $-1 < r < 1$,we can find the range for $b$:
If $r \to 1$,then $b \to 5(1 - 1) = 0$.
If $r \to -1$,then $b \to 5(1 - (-1)) = 5(2) = 10$.
Thus,for $-1 < r < 1$,the value of $b$ lies in the interval $(0, 10)$.

(D) The given series is $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \dots$
The $n$-th term $T_n$ can be written as $T_n = \frac{2^n - 1}{2^{n-1}} = 2 - \frac{1}{2^{n-1}}$ for $n \ge 1$.
The sum of the first $20$ terms is $S_{20} = \sum_{n=1}^{20} (2 - \frac{1}{2^{n-1}})$.
$S_{20} = \sum_{n=1}^{20} 2 - \sum_{n=1}^{20} \frac{1}{2^{n-1}}$.
$S_{20} = 2(20) - (1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{19}})$.
The second part is a geometric progression with $a = 1$,$r = \frac{1}{2}$,and $n = 20$.
Sum $= \frac{1(1 - (1/2)^{20})}{1 - 1/2} = 2(1 - \frac{1}{2^{20}}) = 2 - \frac{1}{2^{19}}$.
Therefore,$S_{20} = 40 - (2 - \frac{1}{2^{19}}) = 38 + \frac{1}{2^{19}}$.

(D) We need to evaluate the sum $S = \sum\limits_{r = 16}^{30} {(r^2 - r - 6)}$.
Using the summation properties,$S = \sum\limits_{r = 16}^{30} r^2 - \sum\limits_{r = 16}^{30} r - \sum\limits_{r = 16}^{30} 6$.
Recall that $\sum\limits_{r = 1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum\limits_{r = 1}^{n} r = \frac{n(n+1)}{2}$.
$\sum\limits_{r = 16}^{30} r^2 = \sum\limits_{r = 1}^{30} r^2 - \sum\limits_{r = 1}^{15} r^2 = \frac{30(31)(61)}{6} - \frac{15(16)(31)}{6} = 9455 - 1240 = 8215$.
$\sum\limits_{r = 16}^{30} r = \sum\limits_{r = 1}^{30} r - \sum\limits_{r = 1}^{15} r = \frac{30(31)}{2} - \frac{15(16)}{2} = 465 - 120 = 345$.
$\sum\limits_{r = 16}^{30} 6 = 6 \times (30 - 16 + 1) = 6 \times 15 = 90$.
Therefore,$S = 8215 - 345 - 90 = 7780$.

(D) Given $f(n) = \left[ \frac{1}{3} + \frac{3n}{100} \right]n$.
For $1 \le n \le 22$,$\frac{1}{3} + \frac{3n}{100} < \frac{1}{3} + \frac{66}{100} = 0.333 + 0.66 = 0.993 < 1$. Thus,$\left[ \frac{1}{3} + \frac{3n}{100} \right] = 0$,so $f(n) = 0$.
For $23 \le n \le 55$,$1 \le \frac{1}{3} + \frac{3n}{100} < \frac{1}{3} + \frac{165}{100} = 0.333 + 1.65 = 1.983 < 2$. Thus,$\left[ \frac{1}{3} + \frac{3n}{100} \right] = 1$,so $f(n) = n$.
For $n = 56$,$\frac{1}{3} + \frac{3(56)}{100} = 0.333 + 1.68 = 2.013$. Thus,$\left[ \frac{1}{3} + \frac{3(56)}{100} \right] = 2$,so $f(56) = 2 \times 56 = 112$.
The sum is $\sum_{n=1}^{56} f(n) = \sum_{n=1}^{22} 0 + \sum_{n=23}^{55} n + f(56)$.
$= 0 + \frac{(55-23+1)}{2}(23+55) + 112 = \frac{33}{2}(78) + 112 = 33 \times 39 + 112 = 1287 + 112 = 1399$.

(C) The $n^{th}$ term $T_n$ of the series is given by:
$T_n = \frac{2n + 1}{\sum_{k=1}^{n} k^2} = \frac{2n + 1}{\frac{n(n+1)(2n+1)}{6}} = \frac{6}{n(n+1)}$
Using partial fractions,we can write:
$T_n = 6 \left( \frac{1}{n} - \frac{1}{n+1} \right)$
Now,the sum of $n$ terms $S_n$ is:
$S_n = \sum_{k=1}^{n} T_k = 6 \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right)$
$S_n = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) \right]$
$S_n = 6 \left( 1 - \frac{1}{n+1} \right) = \frac{6n}{n+1}$
For $n = 11$ terms:
$S_{11} = \frac{6 \times 11}{11 + 1} = \frac{66}{12} = \frac{11}{2}$

(C) The given series is $S = 1(2)^2 + 2(4)^2 + 3(6)^2 + \dots + 10(20)^2$.
The $n^{th}$ term is $T_n = n(2n)^2 = n(4n^2) = 4n^3$.
The sum of $10$ terms is $S_{10} = \sum_{n=1}^{10} 4n^3$.
$S_{10} = 4 \sum_{n=1}^{10} n^3$.
Using the formula $\sum_{n=1}^{k} n^3 = \left( \frac{k(k+1)}{2} \right)^2$,we get:
$S_{10} = 4 \left( \frac{10 \times 11}{2} \right)^2$.
$S_{10} = 4 \times (55)^2 = 4 \times 3025 = 12100$.

(A) The given series is the sum of squares of the first $13$ odd numbers.
The general term is ${T_n} = {(2n - 1)^2}$ for $n = 1, 2, \dots, 13$.
The sum is $S = \sum_{n=1}^{13} (2n - 1)^2 = \sum_{n=1}^{13} (4n^2 - 4n + 1)$.
Using the summation formulas $\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$ and $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$:
$S = 4 \sum_{n=1}^{13} n^2 - 4 \sum_{n=1}^{13} n + \sum_{n=1}^{13} 1$
$S = 4 \left[ \frac{13(13+1)(2 \times 13 + 1)}{6} \right] - 4 \left[ \frac{13(13+1)}{2} \right] + 13$
$S = 4 \left[ \frac{13 \times 14 \times 27}{6} \right] - 2 \times 13 \times 14 + 13$
$S = 4 \times 13 \times 7 \times 9 - 364 + 13$
$S = 3276 - 364 + 13 = 2925$.

(A) The given series is $S = (1^2 + 3^2 + 5^2 + \dots + (2m-1)^2) + 2(2^2 + 4^2 + 6^2 + \dots + (2m)^2)$.
Sum of odd squares up to $(2m-1)^2$ is $\sum_{k=1}^{m} (2k-1)^2 = \frac{m(2m-1)(2m+1)}{3}$.
Sum of even squares up to $(2m)^2$ is $2 \times \sum_{k=1}^{m} (2k)^2 = 8 \sum_{k=1}^{m} k^2 = 8 \times \frac{m(m+1)(2m+1)}{6} = \frac{4m(m+1)(2m+1)}{3}$.
Adding these,$S = \frac{m(2m+1)}{3} [ (2m-1) + 4(m+1) ] = \frac{m(2m+1)}{3} [ 2m - 1 + 4m + 4 ] = \frac{m(2m+1)(6m+3)}{3} = m(2m+1)(2m+1) = m(2m+1)^2$.

(C) The given series is $S_n = 1 + \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + \dots + n \text{ terms}$.
We can rewrite the terms as:
$S_n = (1) + (1 + \frac{1}{3}) + (1 + \frac{1}{9}) + (1 + \frac{1}{27}) + \dots + n \text{ terms}$.
Grouping the terms,we get:
$S_n = (1 + 1 + 1 + \dots + n \text{ times}) + (\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots + n \text{ terms})$.
$S_n = n + \frac{\frac{1}{3}(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}}$.
$S_n = n + \frac{\frac{1}{3}(1 - \frac{1}{3^n})}{\frac{2}{3}}$.
$S_n = n + \frac{1}{2}(1 - \frac{1}{3^n})$.
$S_n = n + \frac{1}{2} - \frac{1}{2 \cdot 3^n}$.

(C) For an odd $n$,the series consists of $n$ terms: $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \dots + 2 \cdot (n-1)^2 + n^2$.
This can be written as the sum of the first $(n-1)$ terms plus the $n$-th term.
Since $(n-1)$ is even,we use the given formula for even terms: $S_{n-1} = \frac{(n-1)(n-1+1)^2}{2} = \frac{(n-1)n^2}{2}$.
Now,adding the $n$-th term $(n^2)$: $S_n = \frac{(n-1)n^2}{2} + n^2$.
Factoring out $n^2$: $S_n = n^2 \left( \frac{n-1}{2} + 1 \right) = n^2 \left( \frac{n-1+2}{2} \right) = \frac{n^2(n+1)}{2}$.

(A) The $n^{th}$ term of the series is given by $T_n = \frac{(3 + (n-1) \times 3)(1^2 + 2^2 + ... + n^2)}{2n + 1}$.
Simplifying $T_n$:
$T_n = \frac{3n \times \frac{n(n+1)(2n+1)}{6}}{2n+1} = \frac{n^2(n+1)}{2} = \frac{n^3 + n^2}{2}$.
The sum of $15$ terms is $S_{15} = \sum_{n=1}^{15} T_n = \frac{1}{2} \sum_{n=1}^{15} (n^3 + n^2)$.
Using the summation formulas $\sum n^3 = [\frac{n(n+1)}{2}]^2$ and $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$:
$S_{15} = \frac{1}{2} [(\frac{15 \times 16}{2})^2 + \frac{15 \times 16 \times 31}{6}]$.
$S_{15} = \frac{1}{2} [120^2 + 1240] = \frac{1}{2} [14400 + 1240] = \frac{15640}{2} = 7820$.

(C) Given $S_k = \frac{k(k+1)}{2k} = \frac{k+1}{2}$.
We need to find $A$ such that $\sum_{k=1}^{10} S_k^2 = \frac{5}{12}A$.
$\sum_{k=1}^{10} \left( \frac{k+1}{2} \right)^2 = \frac{5}{12}A$
$\frac{1}{4} \sum_{k=1}^{10} (k+1)^2 = \frac{5}{12}A$
$\frac{1}{4} (2^2 + 3^2 + .... + 11^2) = \frac{5}{12}A$
We know $\sum_{n=1}^{n} n^2 = \frac{n(n+1)(2n+1)}{6}$.
So,$\sum_{k=1}^{11} k^2 = \frac{11(12)(23)}{6} = 11 \times 2 \times 23 = 506$.
Therefore,$2^2 + 3^2 + .... + 11^2 = 506 - 1^2 = 505$.
Substituting this back: $\frac{1}{4} (505) = \frac{5}{12}A$.
$A = 505 \times \frac{12}{4 \times 5} = 505 \times \frac{3}{5} = 101 \times 3 = 303$.

(B) The given series is ${\left( {\frac{3}{4}} \right)^3} + {\left( {\frac{6}{4}} \right)^3} + {\left( {\frac{9}{4}} \right)^3} + {\left( {\frac{12}{4}} \right)^3} + \dots$ up to $15$ terms.
This can be written as $\sum_{r=1}^{15} {\left( \frac{3r}{4} \right)^3}$.
$= \frac{27}{64} \sum_{r=1}^{15} r^3$.
Using the formula $\sum_{r=1}^{n} r^3 = {\left[ \frac{n(n+1)}{2} \right]^2}$,we get:
$= \frac{27}{64} \times {\left[ \frac{15(16)}{2} \right]^2}$.
$= \frac{27}{64} \times (120)^2$.
$= \frac{27}{64} \times 14400$.
$= 27 \times 225$.
Given that the sum is $225\,k$,we have $225\,k = 225 \times 27$.
Therefore,$k = 27$.

(B) The $n^{th}$ term of the series is given by $T_n = n(2n - 1) = 2n^2 - n$.
The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} (2k^2 - k) = 2 \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k$.
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:
$S_n = 2 \left[ \frac{n(n+1)(2n+1)}{6} \right] - \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{3} - \frac{n(n+1)}{2}$.
For $n = 11$:
$S_{11} = \frac{11(12)(23)}{3} - \frac{11(12)}{2}$.
$S_{11} = 11 \times 4 \times 23 - 11 \times 6$.
$S_{11} = 1012 - 66 = 946$.

(B) The $n^{th}$ term $T_n$ is given by:
$T_n = \frac{(2n+1) \sum_{k=1}^n k^3}{\sum_{k=1}^n k^2}$
Using the formulas $\sum k^3 = \frac{n^2(n+1)^2}{4}$ and $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$:
$T_n = \frac{(2n+1) \times \frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(2n+1)}{6}}$
$T_n = \frac{6}{4} n(n+1) = \frac{3}{2}(n^2 + n)$
Now,find the sum $S_{10} = \sum_{n=1}^{10} T_n = \frac{3}{2} \left[ \sum_{n=1}^{10} n^2 + \sum_{n=1}^{10} n \right]$
$S_{10} = \frac{3}{2} \left[ \frac{10(11)(21)}{6} + \frac{10(11)}{2} \right]$
$S_{10} = \frac{3}{2} [385 + 55] = \frac{3}{2} [440] = 660$

(A) Let the given sum be $S$. The $n^{th}$ term of the series is $T_n = \frac{\sum_{k=1}^n k^3}{\sum_{k=1}^n k}$.
Using the formulas $\sum_{k=1}^n k^3 = \left[\frac{n(n+1)}{2}\right]^2$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,we get $T_n = \frac{[n(n+1)/2]^2}{n(n+1)/2} = \frac{n(n+1)}{2}$.
The sum of the first $15$ terms is $\sum_{n=1}^{15} T_n = \sum_{n=1}^{15} \frac{n(n+1)}{2} = \frac{1}{2} \left[ \sum_{n=1}^{15} n^2 + \sum_{n=1}^{15} n \right]$.
Using $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$,for $n=15$:
Sum $= \frac{1}{2} \left[ \frac{15 \times 16 \times 31}{6} + \frac{15 \times 16}{2} \right] = \frac{1}{2} [1240 + 120] = \frac{1360}{2} = 680$.
The expression is $\sum_{n=1}^{15} T_n - \frac{1}{2} \sum_{n=1}^{15} n = 680 - \frac{1}{2} \times 120 = 680 - 60 = 620$.

(C) Let $S = \sum_{k=0}^{99} \left[ -\frac{1}{3} - \frac{k}{100} \right]$.
We know that $[x] = -1$ if $-1 \le x < 0$ and $[x] = -2$ if $-2 \le x < -1$.
For the term $\left[ -\frac{1}{3} - \frac{k}{100} \right]$,the value is $-1$ when $-\frac{1}{3} - \frac{k}{100} \ge -1$,which implies $\frac{k}{100} \le 1 - \frac{1}{3} = \frac{2}{3}$,so $k \le \frac{200}{3} \approx 66.66$.
Thus,for $k = 0, 1, 2, \dots, 66$ (total $67$ terms),the value is $-1$.
For $k = 67, 68, \dots, 99$ (total $99 - 67 + 1 = 33$ terms),the value is $-2$.
Sum $= 67 \times (-1) + 33 \times (-2) = -67 - 66 = -133$.