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

(C) We know that $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$,$\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$,and $\sum_{n=1}^{k} n^3 = \left(\frac{k(k+1)}{2}\right)^2$.
Given the sum is $\frac{1}{4} \sum_{n=1}^{7} (2n^3 + 3n^2 + n)$.
Using the formulas for $k=7$:
$\sum_{n=1}^{7} n = \frac{7 \times 8}{2} = 28$.
$\sum_{n=1}^{7} n^2 = \frac{7 \times 8 \times 15}{6} = 140$.
$\sum_{n=1}^{7} n^3 = (28)^2 = 784$.
Substituting these values:
Sum $= \frac{1}{4} [2(784) + 3(140) + 28] = \frac{1}{4} [1568 + 420 + 28] = \frac{2016}{4} = 504$.

(C) The sum is given by $\sum_{k=1}^{20} \frac{k(k+1)}{2}$.
Using the formula for the sum of the first $n$ natural numbers,$1+2+\ldots+k = \frac{k(k+1)}{2}$.
We need to calculate $\frac{1}{2} \sum_{k=1}^{20} (k^2 + k)$.
Using the summation formulas $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ for $n=20$:
$\sum_{k=1}^{20} k^2 = \frac{20(21)(41)}{6} = 2870$.
$\sum_{k=1}^{20} k = \frac{20(21)}{2} = 210$.
Therefore,the total sum is $\frac{1}{2} (2870 + 210) = \frac{1}{2} (3080) = 1540$.

(A) To find the $20^{\text{th}}$ term,we substitute $n = 20$ into the given formula:
$a_{20} = (20-1)(2-20)(3+20)$
$a_{20} = (19) \times (-18) \times (23)$
$a_{20} = -342 \times 23$
$a_{20} = -7866$

(A) Given the $n^{th}$ term is $a_{n} = n(n+2)$.
Substituting $n = 1, 2, 3, 4, 5$:
For $n = 1: a_{1} = 1(1+2) = 3$
For $n = 2: a_{2} = 2(2+2) = 8$
For $n = 3: a_{3} = 3(3+2) = 15$
For $n = 4: a_{4} = 4(4+2) = 24$
For $n = 5: a_{5} = 5(5+2) = 35$
Thus,the first five terms are $3, 8, 15, 24, 35$.

(A) Given the $n^{th}$ term $a_{n} = \frac{n}{n+1}$.
Substituting $n = 1, 2, 3, 4, 5$,we obtain:
$a_{1} = \frac{1}{1+1} = \frac{1}{2}$
$a_{2} = \frac{2}{2+1} = \frac{2}{3}$
$a_{3} = \frac{3}{3+1} = \frac{3}{4}$
$a_{4} = \frac{4}{4+1} = \frac{4}{5}$
$a_{5} = \frac{5}{5+1} = \frac{5}{6}$
Therefore,the first five terms are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}$.

(A) To find the first five terms,we substitute $n = 1, 2, 3, 4, 5$ into the formula $a_{n} = n \frac{n^{2}+5}{4}$.
For $n=1$: $a_{1} = 1 \cdot \frac{1^{2}+5}{4} = \frac{6}{4} = \frac{3}{2}$.
For $n=2$: $a_{2} = 2 \cdot \frac{2^{2}+5}{4} = 2 \cdot \frac{9}{4} = \frac{9}{2}$.
For $n=3$: $a_{3} = 3 \cdot \frac{3^{2}+5}{4} = 3 \cdot \frac{14}{4} = \frac{42}{4} = \frac{21}{2}$.
For $n=4$: $a_{4} = 4 \cdot \frac{4^{2}+5}{4} = 4 \cdot \frac{21}{4} = 21$.
For $n=5$: $a_{5} = 5 \cdot \frac{5^{2}+5}{4} = 5 \cdot \frac{30}{4} = \frac{150}{4} = \frac{75}{2}$.
Thus,the first five terms are $\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21, \frac{75}{2}$.

(A) The $n^{\text{th}}$ term of the sequence is given by $a_{n} = (-1)^{n-1} n^{3}$.
To find the $9^{\text{th}}$ term,we substitute $n = 9$ into the formula:
$a_{9} = (-1)^{9-1} (9)^{3}$
$a_{9} = (-1)^{8} (729)$
Since $(-1)^{8} = 1$,we have:
$a_{9} = 1 \times 729 = 729$.

(A) To find the $20^{\text{th}}$ term,we substitute $n = 20$ into the given formula for the $n^{\text{th}}$ term:
$a_{n} = \frac{n(n-2)}{n+3}$
Substituting $n = 20$:
$a_{20} = \frac{20(20-2)}{20+3}$
$a_{20} = \frac{20(18)}{23}$
$a_{20} = \frac{360}{23}$

(A) Given $a_1 = 1, a_2 = 1$ and $a_n = a_{n-1} + a_{n-2}$ for $n > 2$.
Calculating the first few terms:
$a_3 = a_2 + a_1 = 1 + 1 = 2$
$a_4 = a_3 + a_2 = 2 + 1 = 3$
$a_5 = a_4 + a_3 = 3 + 2 = 5$
$a_6 = a_5 + a_4 = 5 + 3 = 8$
Now,calculating the ratios $\frac{a_{n+1}}{a_n}$ for $n = 1, 2, 3, 4, 5$:
For $n = 1: \frac{a_2}{a_1} = \frac{1}{1} = 1$
For $n = 2: \frac{a_3}{a_2} = \frac{2}{1} = 2$
For $n = 3: \frac{a_4}{a_3} = \frac{3}{2}$
For $n = 4: \frac{a_5}{a_4} = \frac{5}{3}$
For $n = 5: \frac{a_6}{a_5} = \frac{8}{5}$
Thus,the sequence of ratios is $1, 2, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}$.

(A) Let the sum be $S_n = 7 + 77 + 777 + 7777 + \ldots$ to $n$ terms.
We can write this as:
$S_n = \frac{7}{9} [9 + 99 + 999 + 9999 + \ldots \text{ to } n \text{ terms}]$
$S_n = \frac{7}{9} [(10 - 1) + (10^2 - 1) + (10^3 - 1) + \ldots + (10^n - 1)]$
$S_n = \frac{7}{9} [(10 + 10^2 + 10^3 + \ldots + 10^n) - (1 + 1 + 1 + \ldots + 1 \text{ to } n \text{ terms})]$
The first part is a geometric progression with first term $a = 10$ and common ratio $r = 10$.
Using the sum formula $S = \frac{a(r^n - 1)}{r - 1}$:
$S_n = \frac{7}{9} [\frac{10(10^n - 1)}{10 - 1} - n]$
$S_n = \frac{7}{9} [\frac{10(10^n - 1)}{9} - n]$
$S_n = \frac{7}{81} [10(10^n - 1) - 9n]$

(B) The given sequence is $8, 88, 888, 8888, \ldots$ to $n$ terms.
Let $S_n = 8 + 88 + 888 + 8888 + \ldots$ to $n$ terms.
$S_n = 8(1 + 11 + 111 + 1111 + \ldots \text{ to } n \text{ terms})$.
$S_n = \frac{8}{9}(9 + 99 + 999 + 9999 + \ldots \text{ to } n \text{ terms})$.
$S_n = \frac{8}{9}[(10-1) + (10^2-1) + (10^3-1) + \ldots + (10^n-1)]$.
$S_n = \frac{8}{9}[(10 + 10^2 + 10^3 + \ldots + 10^n) - (1 + 1 + 1 + \ldots + 1 \text{ to } n \text{ terms})]$.
The sum of the geometric series $10 + 10^2 + \ldots + 10^n$ is $\frac{10(10^n-1)}{10-1} = \frac{10}{9}(10^n-1)$.
Thus,$S_n = \frac{8}{9} [\frac{10}{9}(10^n-1) - n]$.
$S_n = \frac{80}{81}(10^n-1) - \frac{8}{9}n$.

(D) Let the $n^{th}$ term be $a_n$. The series is $5, 11, 19, 29, 41, \ldots$
The differences between consecutive terms are $6, 8, 10, 12, \ldots$,which form an arithmetic progression.
Thus,$a_n = An^2 + Bn + C$.
For $n=1, A+B+C=5$.
For $n=2, 4A+2B+C=11$.
For $n=3, 9A+3B+C=19$.
Subtracting the equations,we get $3A+B=6$ and $5A+B=8$. Solving these,$2A=2 \implies A=1$,$B=3$,and $C=1$.
So,$a_n = n^2 + 3n + 1$.
Now,$S_n = \sum_{k=1}^n (k^2 + 3k + 1) = \sum_{k=1}^n k^2 + 3\sum_{k=1}^n k + \sum_{k=1}^n 1$.
$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} + n$.
$S_n = \frac{n(2n^2 + 3n + 1) + 9n(n+1) + 6n}{6} = \frac{2n^3 + 3n^2 + n + 9n^2 + 9n + 6n}{6} = \frac{2n^3 + 12n^2 + 16n}{6} = \frac{n(n^2 + 6n + 8)}{3} = \frac{n(n+2)(n+4)}{3}$.

(B) Given that the $n^{\text{th}}$ term is $a_n = n(n+3) = n^2 + 3n$.
The sum to $n$ terms is given by $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (k^2 + 3k)$.
Using the standard summation formulas $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,we get:
$S_n = \frac{n(n+1)(2n+1)}{6} + 3 \times \frac{n(n+1)}{2}$.
Taking $\frac{n(n+1)}{2}$ as a common factor:
$S_n = \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 3 \right] = \frac{n(n+1)}{2} \left[ \frac{2n+1+9}{3} \right] = \frac{n(n+1)(2n+10)}{6}$.
$S_n = \frac{n(n+1) \times 2(n+5)}{6} = \frac{n(n+1)(n+5)}{3}$.

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

(A) The $n^{th}$ term of the series is given by $a_{n} = (2n+1)n^{2} = 2n^{3} + n^{2}$.
Sum to $n$ terms $S_{n} = \sum_{k=1}^{n} a_{k} = \sum_{k=1}^{n} (2k^{3} + k^{2})$.
$S_{n} = 2 \sum_{k=1}^{n} k^{3} + \sum_{k=1}^{n} k^{2}$.
Using the formulas $\sum k^{3} = [\frac{n(n+1)}{2}]^{2}$ and $\sum k^{2} = \frac{n(n+1)(2n+1)}{6}$:
$S_{n} = 2 [\frac{n(n+1)}{2}]^{2} + \frac{n(n+1)(2n+1)}{6}$.
$S_{n} = \frac{n^{2}(n+1)^{2}}{2} + \frac{n(n+1)(2n+1)}{6}$.
$S_{n} = \frac{n(n+1)}{2} [n(n+1) + \frac{2n+1}{3}]$.
$S_{n} = \frac{n(n+1)}{2} [\frac{3n^{2} + 3n + 2n + 1}{3}]$.
$S_{n} = \frac{n(n+1)(3n^{2} + 5n + 1)}{6}$.

(A) The given series is $5^{2} + 6^{2} + 7^{2} + \ldots + 20^{2}$.
This can be written as the difference of two sums of squares: $\sum_{k=1}^{20} k^{2} - \sum_{k=1}^{4} k^{2}$.
Using the formula $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$:
For $n=20$: $\sum_{k=1}^{20} k^{2} = \frac{20(21)(41)}{6} = 10 \times 7 \times 41 = 2870$.
For $n=4$: $\sum_{k=1}^{4} k^{2} = \frac{4(5)(9)}{6} = 2 \times 5 \times 3 = 30$.
Therefore,the sum is $2870 - 30 = 2840$.

(C) The $n^{\text{th}}$ term $a_n$ of the series is given by the product of the $n^{\text{th}}$ terms of the two sequences $(3, 6, 9, \dots)$ and $(8, 11, 14, \dots)$.
$a_n = (3n) \times (3n + 5) = 9n^2 + 15n$.
To find the sum $S_n = \sum_{k=1}^n a_k$,we calculate:
$S_n = \sum_{k=1}^n (9k^2 + 15k) = 9 \sum_{k=1}^n k^2 + 15 \sum_{k=1}^n k$.
Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$:
$S_n = 9 \times \frac{n(n+1)(2n+1)}{6} + 15 \times \frac{n(n+1)}{2}$.
$S_n = \frac{3n(n+1)(2n+1)}{2} + \frac{15n(n+1)}{2}$.
$S_n = \frac{3n(n+1)}{2} (2n + 1 + 5) = \frac{3n(n+1)}{2} (2n + 6)$.
$S_n = 3n(n+1)(n+3)$.

(A) The $k$-th term of the series is $a_{k} = \sum_{i=1}^{k} i^{2} = \frac{k(k+1)(2k+1)}{6}$.
Expanding this,we get $a_{k} = \frac{2k^{3} + 3k^{2} + k}{6} = \frac{1}{3}k^{3} + \frac{1}{2}k^{2} + \frac{1}{6}k$.
The sum to $n$ terms is $S_{n} = \sum_{k=1}^{n} a_{k} = \sum_{k=1}^{n} (\frac{1}{3}k^{3} + \frac{1}{2}k^{2} + \frac{1}{6}k)$.
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 $\sum k = \frac{n(n+1)}{2}$:
$S_{n} = \frac{1}{3} \cdot \frac{n^{2}(n+1)^{2}}{4} + \frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{1}{6} \cdot \frac{n(n+1)}{2}$.
Factor out $\frac{n(n+1)}{12}$:
$S_{n} = \frac{n(n+1)}{12} [n(n+1) + (2n+1) + 1] = \frac{n(n+1)}{12} [n^{2} + n + 2n + 2] = \frac{n(n+1)}{12} [n(n+1) + 2(n+1)]$.
$S_{n} = \frac{n(n+1)(n+1)(n+2)}{12} = \frac{n(n+1)^{2}(n+2)}{12}$.

(A) $a_n = n(n+1)(n+4) = n(n^2 + 5n + 4) = n^3 + 5n^2 + 4n$
$\therefore S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n k^3 + 5\sum_{k=1}^n k^2 + 4\sum_{k=1}^n k$
$= \frac{n^2(n+1)^2}{4} + \frac{5n(n+1)(2n+1)}{6} + \frac{4n(n+1)}{2}$
$= \frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + \frac{5(2n+1)}{3} + 4 \right]$
$= \frac{n(n+1)}{2} \left[ \frac{3n^2 + 3n + 20n + 10 + 24}{6} \right]$
$= \frac{n(n+1)}{2} \left[ \frac{3n^2 + 23n + 34}{6} \right]$
$= \frac{n(n+1)(3n^2 + 23n + 34)}{12}$

(A) Given the $n^{th}$ term is $a_n = n^2 + 2^n$.
The sum of $n$ terms is $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (k^2 + 2^k)$.
This can be split into two separate summations: $S_n = \sum_{k=1}^n k^2 + \sum_{k=1}^n 2^k$.
The sum of the first $n$ squares is given by the formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$.
The second part is a geometric progression: $\sum_{k=1}^n 2^k = 2^1 + 2^2 + \dots + 2^n$.
Using the sum formula for a $G.P.$ with first term $a=2$ and common ratio $r=2$,we get $\frac{a(r^n - 1)}{r-1} = \frac{2(2^n - 1)}{2-1} = 2(2^n - 1) = 2^{n+1} - 2$.
Combining these,$S_n = \frac{n(n+1)(2n+1)}{6} + 2(2^n - 1)$.

(A) The $n^{th}$ term is $a_n = (2n-1)^2 = 4n^2 - 4n + 1$.
The sum to $n$ terms is $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (4k^2 - 4k + 1)$.
Using the summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum k = \frac{n(n+1)}{2}$,and $\sum 1 = n$:
$S_n = 4 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 4 \left[ \frac{n(n+1)}{2} \right] + n$
$S_n = \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n$
$S_n = n \left[ \frac{2(2n^2+3n+1) - 6(n+1) + 3}{3} \right]$
$S_n = n \left[ \frac{4n^2 + 6n + 2 - 6n - 6 + 3}{3} \right]$
$S_n = n \left[ \frac{4n^2 - 1}{3} \right] = \frac{n(2n-1)(2n+1)}{3}$.

(B) Let $S_n = 5 + 55 + 555 + \ldots$ to $n$ terms.
$S_n = 5(1 + 11 + 111 + \ldots \text{ to } n \text{ terms})$
$S_n = \frac{5}{9}(9 + 99 + 999 + \ldots \text{ to } n \text{ terms})$
$S_n = \frac{5}{9}((10-1) + (10^2-1) + (10^3-1) + \ldots + (10^n-1))$
$S_n = \frac{5}{9}((10 + 10^2 + 10^3 + \ldots + 10^n) - (1 + 1 + 1 + \ldots + 1))$
Using the sum formula for a geometric progression $a(r^n-1)/(r-1)$ where $a=10$ and $r=10$:
$S_n = \frac{5}{9}\left(\frac{10(10^n-1)}{10-1} - n\right)$
$S_n = \frac{5}{9}\left(\frac{10(10^n-1)}{9} - n\right)$
$S_n = \frac{50}{81}(10^n-1) - \frac{5n}{9}$

(A) Let $S_n = 0.6 + 0.66 + 0.666 + \dots$ to $n$ terms.
$S_n = 6 [0.1 + 0.11 + 0.111 + \dots \text{ to } n \text{ terms}]$
$S_n = \frac{6}{9} [0.9 + 0.99 + 0.999 + \dots \text{ to } n \text{ terms}]$
$S_n = \frac{2}{3} [(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \dots \text{ to } n \text{ terms}]$
$S_n = \frac{2}{3} [n - (0.1 + 0.01 + 0.001 + \dots \text{ to } n \text{ terms})]$
The term in the bracket is a geometric progression with $a = 0.1$ and $r = 0.1$.
$S_n = \frac{2}{3} [n - \frac{0.1(1 - (0.1)^n)}{1 - 0.1}]$
$S_n = \frac{2}{3} [n - \frac{0.1(1 - 10^{-n})}{0.9}]$
$S_n = \frac{2}{3} [n - \frac{1}{9}(1 - 10^{-n})]$
$S_n = \frac{2}{3} n - \frac{2}{27}(1 - 10^{-n})$

(A) The given series is $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$ up to $n$ terms.
The $n^{\text{th}}$ term of the series is given by $a_n = (2n) \times (2n + 2)$.
$a_n = 4n^2 + 4n$.
To find the $20^{\text{th}}$ term,substitute $n = 20$ into the formula:
$a_{20} = 4(20)^2 + 4(20)$.
$a_{20} = 4(400) + 80$.
$a_{20} = 1600 + 80 = 1680$.
Thus,the $20^{\text{th}}$ term of the series is $1680$.

(A) The given series is $3+7+13+21+31+\ldots$
Let $a_n$ be the $n^{th}$ term of the series.
The differences between consecutive terms are $4, 6, 8, 10, \ldots$,which form an arithmetic progression.
Thus,$a_n = a_1 + \sum_{k=1}^{n-1} (4 + (k-1)2) = 3 + \sum_{k=1}^{n-1} (2k+2) = 3 + 2\frac{(n-1)n}{2} + 2(n-1) = 3 + n^2 - n + 2n - 2 = n^2 + n + 1$.
Now,the sum of the first $n$ terms is $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (k^2 + k + 1)$.
$S_n = \sum_{k=1}^n k^2 + \sum_{k=1}^n k + \sum_{k=1}^n 1$.
$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + n$.
$S_n = \frac{n}{6} [ (n+1)(2n+1) + 3(n+1) + 6 ]$.
$S_n = \frac{n}{6} [ 2n^2 + 3n + 1 + 3n + 3 + 6 ]$.
$S_n = \frac{n}{6} [ 2n^2 + 6n + 10 ]$.
$S_n = \frac{n}{3} [ n^2 + 3n + 5 ]$.

(N/A) Given that:
$S_{1} = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
$S_{2} = \sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$
$S_{3} = \sum_{k=1}^{n} k^{3} = \frac{n^{2}(n+1)^{2}}{4}$
Consider the right-hand side $(RHS)$: $S_{3}(1 + 8 S_{1})$
$= \frac{n^{2}(n+1)^{2}}{4} \left(1 + 8 \cdot \frac{n(n+1)}{2}\right)$
$= \frac{n^{2}(n+1)^{2}}{4} (1 + 4n(n+1))$
$= \frac{n^{2}(n+1)^{2}}{4} (1 + 4n^{2} + 4n)$
$= \frac{n^{2}(n+1)^{2}}{4} (2n+1)^{2}$
$= \frac{[n(n+1)(2n+1)]^{2}}{4} \quad \dots (1)$
Now consider the left-hand side $(LHS)$: $9 S_{2}^{2}$
$= 9 \left( \frac{n(n+1)(2n+1)}{6} \right)^{2}$
$= 9 \cdot \frac{[n(n+1)(2n+1)]^{2}}{36}$
$= \frac{[n(n+1)(2n+1)]^{2}}{4} \quad \dots (2)$
From $(1)$ and $(2)$,we have $9 S_{2}^{2} = S_{3}(1 + 8 S_{1})$.

The $n^{th}$ term of the given series is
$a_n = \frac{1^{3}+2^{3}+3^{3}+\ldots+n^{3}}{1+3+5+\ldots+(2 n-1)} = \frac{[\frac{n(n+1)}{2}]^{2}}{n^2}$
Since $1+3+5+\ldots+(2 n-1)$ is an $A.P.$ with $n$ terms,its sum is $n^2$.
$a_n = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4} = \frac{1}{4}(n^2 + 2n + 1)$
Now,the sum of $n$ terms $S_n = \sum_{k=1}^n a_k = \frac{1}{4} \sum_{k=1}^n (k^2 + 2k + 1)$
$S_n = \frac{1}{4} [\frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2} + n]$
$S_n = \frac{1}{4} [\frac{n(n+1)(2n+1) + 6n(n+1) + 6n}{6}]$
$S_n = \frac{n}{24} [2n^2 + 3n + 1 + 6n + 6 + 6] = \frac{n(2n^2 + 9n + 13)}{24}$

$n^{\text{th}}$ term of the numerator $= n(n+1)^{2} = n^{3}+2n^{2}+n$
$n^{\text{th}}$ term of the denominator $= n^{2}(n+1) = n^{3}+n^{2}$
Let $S_N = \sum_{k=1}^{n} (k^{3}+2k^{2}+k)$ and $S_D = \sum_{k=1}^{n} (k^{3}+k^{2})$.
Using standard summation formulas:
$\sum k^{3} = \frac{n^{2}(n+1)^{2}}{4}$,$\sum k^{2} = \frac{n(n+1)(2n+1)}{6}$,$\sum k = \frac{n(n+1)}{2}$.
$S_N = \frac{n^{2}(n+1)^{2}}{4} + 2 \times \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)}{12} [3n(n+1) + 4(2n+1) + 6] = \frac{n(n+1)}{12} [3n^{2}+11n+10] = \frac{n(n+1)(n+2)(3n+5)}{12}$.
$S_D = \frac{n^{2}(n+1)^{2}}{4} + \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{12} [3n(n+1) + 2(2n+1)] = \frac{n(n+1)}{12} [3n^{2}+7n+2] = \frac{n(n+1)(n+2)(3n+1)}{12}$.
Dividing $S_N$ by $S_D$:
$\frac{S_N}{S_D} = \frac{n(n+1)(n+2)(3n+5)}{n(n+1)(n+2)(3n+1)} = \frac{3n+5}{3n+1}$.
Thus,the result is proved.

(B) The given expression is $S = 1 + \sum_{n=1}^{10} (1 - (2n)^2(2n-1))$.
This can be written as $S = 1 + \sum_{n=1}^{10} 1 - \sum_{n=1}^{10} (4n^2)(2n-1)$.
$S = 1 + 10 - 4 \sum_{n=1}^{10} (2n^3 - n^2)$.
$S = 11 - 4 [2 \sum_{n=1}^{10} n^3 - \sum_{n=1}^{10} n^2]$.
Using the formulas $\sum n^3 = [\frac{n(n+1)}{2}]^2$ and $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$:
$S = 11 - 4 [2 \cdot (55)^2 - \frac{10 \cdot 11 \cdot 21}{6}]$.
$S = 11 - 4 [2 \cdot 3025 - 385] = 11 - 4 [6050 - 385] = 11 - 4 [5665]$.
$S = 11 - 22660 = 11 - 220(103)$.
Comparing this with $\alpha - 220 \beta$,we get $\alpha = 11$ and $\beta = 103$.
Thus,the ordered pair is $(11, 103)$.

(B) Let the sum be $S = \sum_{k=0}^{99} (100-k)(100+k)$.
$S = \sum_{k=0}^{99} (100^2 - k^2) = \sum_{k=0}^{99} 100^2 - \sum_{k=0}^{99} k^2$.
Since there are $100$ terms (from $k=0$ to $99$):
$S = 100(100^2) - \frac{99(99+1)(2 \times 99 + 1)}{6} = 100^3 - \frac{99 \times 100 \times 199}{6}$.
$S = 1000000 - 33 \times 50 \times 199 = 1000000 - 328350 = 671650$.
Given $100^{\alpha} - 199\beta = 671650$.
If $\alpha = 3$,then $100^3 - 199\beta = 671650 \implies 1000000 - 671650 = 199\beta$.
$328350 = 199\beta \implies \beta = \frac{328350}{199} = 1650$.
Thus,the point is $(3, 1650)$.
The slope of the line passing through $(3, 1650)$ and $(0, 0)$ is $m = \frac{1650 - 0}{3 - 0} = 550$.

(C) We express the term inside the summation as:
$r^3 + 6r^2 + 2r + 5 = (r+1)(r+2)(r+3) - 9(r+1) + 8$.
Thus,the sum becomes:
$\sum_{r=1}^{10} [r!(r+1)(r+2)(r+3) - 9r!(r+1) + 8r!]$.
Using the property $(r+k)! = r!(r+1)...(r+k)$,we rewrite the terms:
$= \sum_{r=1}^{10} [(r+3)! - 9(r+1)! + 8r!]$.
$= \sum_{r=1}^{10} [(r+3)! - (r+1)! - 8((r+1)! - r!)]$.
Expanding the summation:
$= [(4! - 2!) + (5! - 3!) + ... + (13! - 11!)] - 8[(2! - 1!) + (3! - 2!) + ... + (11! - 10!)]$.
$= (13! + 12! - 2! - 3!) - 8(11! - 1!)$.
$= (13 \times 12 \times 11! + 12 \times 11! - 2 - 6) - 8(11!) + 8$.
$= (156 + 12 - 8) \times 11! - 8 + 8$.
$= 160 \times 11!$.
Comparing with $\alpha(11!)$,we get $\alpha = 160$.

(D) Given $x = \sum_{n=0}^{\infty} (\cos^2 \theta)^n = \frac{1}{1 - \cos^2 \theta} = \frac{1}{\sin^2 \theta}$ $\Rightarrow \sin^2 \theta = \frac{1}{x}$.
Similarly,$y = \sum_{n=0}^{\infty} (\sin^2 \phi)^n = \frac{1}{1 - \sin^2 \phi} = \frac{1}{\cos^2 \phi}$ $\Rightarrow \cos^2 \phi = \frac{1}{y}$.
And $z = \sum_{n=0}^{\infty} (\cos^2 \theta \sin^2 \phi)^n = \frac{1}{1 - \cos^2 \theta \sin^2 \phi}$ $\Rightarrow 1 - \cos^2 \theta \sin^2 \phi = \frac{1}{z}$.
Since $\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{1}{x} = \frac{x-1}{x}$ and $\sin^2 \phi = 1 - \cos^2 \phi = 1 - \frac{1}{y} = \frac{y-1}{y}$,we have:
$1 - \left(\frac{x-1}{x}\right) \left(\frac{y-1}{y}\right) = \frac{1}{z}$.
$1 - \frac{xy - x - y + 1}{xy} = \frac{1}{z}$ $\Rightarrow \frac{xy - xy + x + y - 1}{xy} = \frac{1}{z}$ $\Rightarrow \frac{x + y - 1}{xy} = \frac{1}{z}$.
$z(x + y - 1) = xy$ $\Rightarrow z(x + y) - z = xy$ $\Rightarrow xy + z = z(x + y)$.

(A) Let the sum be $S = 1 + \frac{2}{3} + \frac{7}{3^{2}} + \frac{12}{3^{3}} + \frac{17}{3^{4}} + \frac{22}{3^{5}} + \ldots$
Multiply by $\frac{1}{3}$: $\frac{S}{3} = \frac{1}{3} + \frac{2}{3^{2}} + \frac{7}{3^{3}} + \frac{12}{3^{4}} + \frac{17}{3^{5}} + \ldots$
Subtracting the two equations: $S - \frac{S}{3} = 1 + (\frac{2}{3} - \frac{1}{3}) + (\frac{7}{3^{2}} - \frac{2}{3^{2}}) + (\frac{12}{3^{3}} - \frac{7}{3^{3}}) + \ldots$
$\frac{2S}{3} = 1 + \frac{1}{3} + \frac{5}{3^{2}} + \frac{5}{3^{3}} + \frac{5}{3^{4}} + \ldots$
$\frac{2S}{3} = 1 + \frac{1}{3} + \frac{\frac{5}{3^{2}}}{1 - \frac{1}{3}} = 1 + \frac{1}{3} + \frac{5}{9} \times \frac{3}{2} = \frac{4}{3} + \frac{5}{6} = \frac{8+5}{6} = \frac{13}{6}$
$S = \frac{13}{6} \times \frac{3}{2} = \frac{13}{4}$

(D) The given sequence is $T_n = (3n+4)(2n+6)$ for $n = 1, 2, \ldots, 10$.
Expanding the expression: $T_n = 6n^2 + 18n + 8n + 24 = 6n^2 + 26n + 24$.
The sum of the first $10$ terms is $S_{10} = \sum_{n=1}^{10} (6n^2 + 26n + 24)$.
Using summation formulas:
$S_{10} = 6 \sum_{n=1}^{10} n^2 + 26 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 24$.
$S_{10} = 6 \left( \frac{10(11)(21)}{6} \right) + 26 \left( \frac{10(11)}{2} \right) + 24(10)$.
$S_{10} = 2310 + 1430 + 240 = 3980$.
The mean is $\frac{S_{10}}{10} = \frac{3980}{10} = 398$.

(B) Let $P = \sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}}$.
Given the recurrence relation $a_{n+2} = 2a_{n+1} + a_{n}$.
Dividing by $8^{n+2}$,we get $\frac{a_{n+2}}{8^{n+2}} = \frac{2a_{n+1}}{8^{n+2}} + \frac{a_{n}}{8^{n+2}}$.
Summing from $n=1$ to $\infty$:
$\sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}} = \frac{2}{8} \sum_{n=1}^{\infty} \frac{a_{n+1}}{8^{n+1}} + \frac{1}{64} \sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}}$.
Let $S = \sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}} = P$. Then $\sum_{n=1}^{\infty} \frac{a_{n+1}}{8^{n+1}} = P - \frac{a_{1}}{8} = P - \frac{1}{8}$.
And $\sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}} = P - \frac{a_{1}}{8} - \frac{a_{2}}{64} = P - \frac{1}{8} - \frac{1}{64}$.
Substituting these into the equation:
$P - \frac{1}{8} - \frac{1}{64} = \frac{1}{4}(P - \frac{1}{8}) + \frac{1}{64}P$.
Multiplying by $64$:
$64P - 8 - 1 = 16(P - \frac{1}{8}) + P$.
$64P - 9 = 16P - 2 + P$.
$64P - 9 = 17P - 2$.
$47P = 7$.

(B) Let $S = \sum_{n=8}^{100} \left[ \frac{(-1)^{n} n}{2} \right]$.
Expanding the sum,we get:
$S = \left[ \frac{8}{2} \right] + \left[ \frac{-9}{2} \right] + \left[ \frac{10}{2} \right] + \left[ \frac{-11}{2} \right] + \dots + \left[ \frac{-99}{2} \right] + \left[ \frac{100}{2} \right]$.
Using the property $[x]$,we have:
$S = 4 + [-4.5] + 5 + [-5.5] + 6 + [-6.5] + \dots + [-49.5] + 50$.
$S = 4 + (-5) + 5 + (-6) + 6 + (-7) + \dots + (-50) + 50$.
Notice that the terms cancel out in pairs: $(-5+5) + (-6+6) + \dots + (-50+50) = 0$.
Therefore,$S = 4 + 0 = 4$.

(B) The $n$-th term of the sequence is $a_n = \frac{3^n - 2^n}{3^n} = 1 - \left(\frac{2}{3}\right)^n$.
The sum of the first $100$ terms is $S_{100} = \sum_{n=1}^{100} \left(1 - \left(\frac{2}{3}\right)^n\right)$.
$S_{100} = \sum_{n=1}^{100} 1 - \sum_{n=1}^{100} \left(\frac{2}{3}\right)^n$.
$S_{100} = 100 - \left[ \frac{2}{3} \frac{(1 - (2/3)^{100})}{1 - 2/3} \right]$.
$S_{100} = 100 - 2 \left(1 - \left(\frac{2}{3}\right)^{100}\right) = 100 - 2 + 2 \left(\frac{2}{3}\right)^{100} = 98 + 2 \left(\frac{2}{3}\right)^{100}$.
Since $0 < 2 \left(\frac{2}{3}\right)^{100} < 1$,the value of $S_{100}$ lies between $98$ and $99$.
Therefore,the greatest integer less than or equal to $S_{100}$ is $[S_{100}] = 98$.

(C) For odd $n$,$3+(-1)^n = 3-1 = 2$. For even $n$,$3+(-1)^n = 3+1 = 4$.
$A = \sum_{k=1}^{\infty} \frac{1}{2^{2k-1}} + \sum_{k=1}^{\infty} \frac{1}{4^{2k}} = (\frac{1}{2} + \frac{1}{2^3} + \dots) + (\frac{1}{4^2} + \frac{1}{4^4} + \dots)$
$A = \frac{1/2}{1-1/4} + \frac{1/16}{1-1/16} = \frac{1/2}{3/4} + \frac{1/16}{15/16} = \frac{2}{3} + \frac{1}{15} = \frac{10+1}{15} = \frac{11}{15}$.
$B = \sum_{k=1}^{\infty} \frac{-1}{2^{2k-1}} + \sum_{k=1}^{\infty} \frac{1}{4^{2k}} = (-\frac{1}{2} - \frac{1}{2^3} - \dots) + (\frac{1}{4^2} + \frac{1}{4^4} + \dots)$
$B = \frac{-1/2}{1-1/4} + \frac{1/16}{1-1/16} = -\frac{2}{3} + \frac{1}{15} = \frac{-10+1}{15} = -\frac{9}{15} = -\frac{3}{5}$.
$\frac{A}{B} = \frac{11/15}{-9/15} = -\frac{11}{9}$.

(D) We know that for any two numbers $i$ and $j$,$\min \{i, j\} + \max \{i, j\} = i + j$.
Therefore,$A + B = \sum_{i=1}^{10} \sum_{j=1}^{10} (\min \{i, j\} + \max \{i, j\}) = \sum_{i=1}^{10} \sum_{j=1}^{10} (i + j)$.
Expanding the sum: $A + B = \sum_{i=1}^{10} (\sum_{j=1}^{10} i + \sum_{j=1}^{10} j) = \sum_{i=1}^{10} (10i + \frac{10 \times 11}{2}) = \sum_{i=1}^{10} (10i + 55)$.
$A + B = 10 \sum_{i=1}^{10} i + \sum_{i=1}^{10} 55 = 10 \times \frac{10 \times 11}{2} + 10 \times 55$.
$A + B = 10 \times 55 + 550 = 550 + 550 = 1100$.

(C) Let $S = 1 + \frac{5}{6} + \frac{12}{6^{2}} + \frac{22}{6^{3}} + \frac{35}{6^{4}} + \ldots$
$\frac{1}{6}S = \frac{1}{6} + \frac{5}{6^{2}} + \frac{12}{6^{3}} + \frac{22}{6^{4}} + \ldots$
Subtracting the two equations:
$\frac{5}{6}S = 1 + \frac{4}{6} + \frac{7}{6^{2}} + \frac{10}{6^{3}} + \frac{13}{6^{4}} + \ldots$
$\frac{5}{36}S = \frac{1}{6} + \frac{4}{6^{2}} + \frac{7}{6^{3}} + \frac{10}{6^{4}} + \ldots$
Subtracting again:
$(\frac{5}{6} - \frac{5}{36})S = 1 + \frac{3}{6} + \frac{3}{6^{2}} + \frac{3}{6^{3}} + \ldots$
$\frac{25}{36}S = 1 + \frac{\frac{3}{6}}{1 - \frac{1}{6}} = 1 + \frac{3}{6} \times \frac{6}{5} = 1 + \frac{3}{5} = \frac{8}{5}$
$S = \frac{8}{5} \times \frac{36}{25} = \frac{288}{125}$

(B) Given the recurrence relation $a_{n+2}-2a_{n+1}+a_{n}=1$ with $a_{0}=0, a_{1}=0$.
Calculating the first few terms: $a_{2}=2(0)-0+1=1$,$a_{3}=2(1)-0+1=3$,$a_{4}=2(3)-1+1=6$.
The general term is $a_{n}=\frac{n(n-1)}{2}$.
Let $S=\sum\limits_{n=2}^{\infty} \frac{n(n-1)}{2 \cdot 7^{n}}$.
This is an arithmetico-geometric series. Using the method of generating functions or series manipulation:
$S = \frac{1}{7^{2}} + \frac{3}{7^{3}} + \frac{6}{7^{4}} + \frac{10}{7^{5}} + \dots$
$\frac{S}{7} = \frac{1}{7^{3}} + \frac{3}{7^{4}} + \frac{6}{7^{5}} + \dots$
Subtracting these: $S(1-\frac{1}{7}) = \frac{1}{7^{2}} + \frac{2}{7^{3}} + \frac{3}{7^{4}} + \dots$
$\frac{6}{7}S = \frac{1}{7^{2}} (1 + \frac{2}{7} + \frac{3}{7^{2}} + \dots) = \frac{1}{49} \cdot \frac{1}{(1-\frac{1}{7})^{2}} = \frac{1}{49} \cdot \frac{1}{(6/7)^{2}} = \frac{1}{49} \cdot \frac{49}{36} = \frac{1}{36}$.
Thus,$S = \frac{1}{36} \cdot \frac{7}{6} = \frac{7}{216}$.

(C) Given $a_{1}=1$ and $a_{n}=a_{n-1}+2$,this is an arithmetic progression with first term $1$ and common difference $2$. Thus,$a_{n} = 1 + (n-1)2 = 2n-1$.
Given $b_{1}=1$ and $b_{n}=a_{n}+b_{n-1}$,we have $b_{n} = (2n-1) + b_{n-1}$.
Calculating the first few terms: $b_{1}=1$,$b_{2}=3+1=4$,$b_{3}=5+4=9$,$b_{4}=7+9=16$. It follows that $b_{n} = n^{2}$.
We need to calculate $\sum_{n=1}^{15} a_{n} b_{n} = \sum_{n=1}^{15} (2n-1)n^{2} = \sum_{n=1}^{15} (2n^{3}-n^{2})$.
Using standard summation formulas:
$\sum_{n=1}^{N} n^{3} = \frac{N^{2}(N+1)^{2}}{4}$ and $\sum_{n=1}^{N} n^{2} = \frac{N(N+1)(2N+1)}{6}$.
For $N=15$:
$2 \times \frac{15^{2} \times 16^{2}}{4} - \frac{15 \times 16 \times 31}{6} = 2 \times \frac{225 \times 256}{4} - \frac{7440}{6} = 28800 - 1240 = 27560$.

(C) Let the $n^{th}$ term be $T_n$. The numerator of the $n^{th}$ term is $S_n = \sum_{k=1}^{2n} (-1)^{k-1} k^3 = 1^3 - 2^3 + 3^3 - 4^3 + \ldots + (2n)^3$.
This can be written as $S_n = \sum_{k=1}^{n} ((2k)^3 - (2k-1)^3)$.
Using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$,we have $(2k)^3 - (2k-1)^3 = (1)(4k^2 + 2k(2k-1) + (2k-1)^2) = 4k^2 + 4k^2 - 2k + 4k^2 - 4k + 1 = 12k^2 - 6k + 1$.
Thus,$S_n = \sum_{k=1}^{n} (12k^2 - 6k + 1) = 12 \frac{n(n+1)(2n+1)}{6} - 6 \frac{n(n+1)}{2} + n = 2n(n+1)(2n+1) - 3n(n+1) + n = n(4n^2 + 6n + 2 - 3n - 3 + 1) = n(4n^2 + 3n) = n^2(4n+3)$.
The denominator of the $n^{th}$ term is $n(4n+3)$.
Therefore,$T_n = \frac{n^2(4n+3)}{n(4n+3)} = n$.
The sum of the series is $\sum_{n=1}^{15} T_n = \sum_{n=1}^{15} n = \frac{15 \times 16}{2} = 120$.

(B) Let $S = \sum_{r=1}^{\infty} \frac{a_{r}}{2^{r}} = \frac{a_{1}}{2} + \frac{a_{2}}{2^{2}} + \frac{a_{3}}{2^{3}} + \ldots = 4$
Multiply by $\frac{1}{2}$:
$\frac{S}{2} = \frac{a_{1}}{2^{2}} + \frac{a_{2}}{2^{3}} + \frac{a_{3}}{2^{4}} + \ldots$
Subtracting the two equations:
$S - \frac{S}{2} = \frac{a_{1}}{2} + \frac{a_{2}-a_{1}}{2^{2}} + \frac{a_{3}-a_{2}}{2^{3}} + \ldots$
Since $a_{r}$ is an $A.P.$,$a_{r} - a_{r-1} = d$:
$\frac{S}{2} = \frac{a_{1}}{2} + \frac{d}{2^{2}} + \frac{d}{2^{3}} + \frac{d}{2^{4}} + \ldots$
$\frac{S}{2} = \frac{a_{1}}{2} + d \left( \frac{1/4}{1 - 1/2} \right) = \frac{a_{1}}{2} + d \left( \frac{1/4}{1/2} \right) = \frac{a_{1}}{2} + \frac{d}{2}$
$S = a_{1} + d = a_{2}$
Given $S = 4$,so $a_{2} = 4$.
Therefore,$4 a_{2} = 4 \times 4 = 16$.

(D) The sequence is defined by $a_n = 20a_{n-1} - 108a_{n-2}$ for $n \geq 2$ with $a_0 = 1, a_1 = 1$.
Multiply by $\frac{1}{10^{2n}}$ and sum from $n=2$ to $\infty$:
$\sum_{n=2}^{\infty} \frac{a_n}{10^{2n}} = \sum_{n=2}^{\infty} \frac{20a_{n-1}}{10^{2n}} - \sum_{n=2}^{\infty} \frac{108a_{n-2}}{10^{2n}}$
$S - a_0 - \frac{a_1}{100} = \frac{20}{100} \sum_{n=2}^{\infty} \frac{a_{n-1}}{10^{2(n-1)}} - \frac{108}{10000} \sum_{n=2}^{\infty} \frac{a_{n-2}}{10^{2(n-2)}}$
$S - 1 - \frac{1}{100} = \frac{1}{5} (S - a_0) - \frac{108}{10000} S$
$S - \frac{101}{100} = \frac{1}{5} (S - 1) - \frac{27}{2500} S$
$S - \frac{1}{5} S + \frac{27}{2500} S = 1 + \frac{1}{100} - \frac{1}{5}$
$S \left( \frac{2500 - 500 + 27}{2500} \right) = \frac{100 + 1 - 20}{100}$
$S \left( \frac{2027}{2500} \right) = \frac{81}{100}$
$S = \frac{81 \times 25}{2027} = \frac{2025}{2027}$
Since $a = 2025$ and $b = 2027$ are coprime,$a = 2025$.

(A) The harmonic series $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$ can be approximated by $\ln(n) + \gamma$,where $\gamma \approx 0.577$ is the Euler-Mascheroni constant.
We want $H_n \geq 4$,so $\ln(n) + 0.577 \approx 4$,which gives $\ln(n) \approx 3.423$.
Calculating $n \approx e^{3.423} \approx 30.66$.
However,using the inequality $\ln(n+1) < 1 + \frac{1}{2} + \dots + \frac{1}{n} < 1 + \ln(n)$:
For $H_n \geq 4$,we have $1 + \ln(n) > 4$,so $\ln(n) > 3$,which means $n > e^3 \approx 20.08$.
More precisely,the value of $n$ for which the harmonic sum reaches $4$ is $n = 31$.
Since $31$ falls in the range $20 < n \leq 60$,option $A$ is correct.

(C) Let $S = 2014^3 - 2013^3 + 2012^3 - 2011^3 + \ldots + 2^3 - 1^3$.
We can group the terms as pairs: $(2014^3 - 2013^3) + (2012^3 - 2011^3) + \ldots + (2^3 - 1^3)$.
Using the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,and noting that $a - b = 1$ for each pair:
$S = (2014^2 + 2014 \times 2013 + 2013^2) + (2012^2 + 2012 \times 2011 + 2011^2) + \ldots + (2^2 + 2 \times 1 + 1^2)$.
There are $1007$ such pairs.
Alternatively,consider the sum $S = \sum_{k=1}^{1007} ((2k)^3 - (2k-1)^3) = \sum_{k=1}^{1007} (8k^3 - (8k^3 - 12k^2 + 6k - 1)) = \sum_{k=1}^{1007} (12k^2 - 6k + 1)$.
$S = 12 \sum_{k=1}^{1007} k^2 - 6 \sum_{k=1}^{1007} k + \sum_{k=1}^{1007} 1$.
Using sum formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$ with $n = 1007$:
$S = 12 \times \frac{1007(1008)(2015)}{6} - 6 \times \frac{1007(1008)}{2} + 1007$.
$S = 2 \times 1007 \times 1008 \times 2015 - 3 \times 1007 \times 1008 + 1007$.
$S = 1007 [2 \times 1008 \times 2015 - 3 \times 1008 + 1]$.
$S = 1007 [4062360 - 3024 + 1] = 1007 \times 4059337$.
Since $4059337 = 4031 \times 1007$,we have $S = 1007^2 \times 4031$.
The largest perfect square dividing $S$ is $1007^2$.

(A) Given $\sum_{k=1}^n (a k^3+b k^2+c k+d)=n^4$.
For $n=1$,$a+b+c+d=1^4=1$.
For $n=2$,$(a+b+c+d) + (8a+4b+2c+d)=2^4=16 \Rightarrow 9a+5b+3c+2d=16$.
For $n=3$,$(9a+5b+3c+2d) + (27a+9b+3c+d)=3^4=81 \Rightarrow 36a+14b+6c+3d=81$.
For $n=4$,$(36a+14b+6c+3d) + (64a+16b+4c+d)=4^4=256 \Rightarrow 100a+30b+10c+4d=256$.
Solving these equations:
$a=4, b=-6, c=4, d=-1$.
Thus,$|a|+|b|+|c|+|d| = |4|+|-6|+|4|+|-1| = 4+6+4+1 = 15$.

(B) Let the general term be $T_r = (r^2 - r + 1)(r!)$.
We can rewrite $T_r$ as:
$T_r = (r^2 + r - 2r + 1)(r!) = (r(r+1) - 2r + 1)(r!)$
Alternatively,observe that $r^2 - r + 1 = r(r-1) + 1$.
$T_r = (r(r-1) + 1)r! = r(r-1)r! + r! = r(r!) (r-1) + r! = r!(r^2 - r + 1)$.
Let us use the method of differences:
$T_r = (r^2 + r - 2r + 1)r! = (r(r+1) - 2r + 1)r! = r(r+1)! - 2r(r!) + r! = r(r+1)! - (2r-1)r!$.
Actually,a simpler approach is:
$T_r = (r^2 + r - 2r + 1)r! = (r+1)!r - (r-1)r!$ is not quite right.
Let $T_r = (r^2 - r + 1)r! = (r^2 + r - 2r + 1)r! = (r(r+1) - 2r + 1)r! = r(r+1)! - 2r(r!) + r! = r(r+1)! - (2r-1)r!$.
Correct approach: $T_r = (r^2 - r + 1)r! = (r^2 + r - 2r + 1)r! = (r(r+1) - 2r + 1)r! = r(r+1)! - 2r(r!) + r! = r(r+1)! - (2r-1)r!$.
Wait,let $T_r = (r^2 + r - 2r + 1)r! = r(r+1)! - 2r(r!) + r! = r(r+1)! - (2r-1)r!$.
Let's test $r=1$: $T_1 = (1-1+1)1! = 1$. Formula: $1(2!) - (1)1! = 2-1 = 1$. Correct.
Let $r=2$: $T_2 = (4-2+1)2! = 3 \times 2 = 6$. Formula: $2(3!) - (3)2! = 12 - 6 = 6$. Correct.
Sum $S_n = \sum_{r=1}^n [r(r+1)! - (2r-1)r!]$.
This is a telescoping sum. $S_n = n(n+1)! - \sum_{r=1}^n (2r-1)r! = n(n+1)! - [1(1!) + 3(2!) + 5(3!) + \ldots + (2n-1)n!]$.
Note that $(2r-1)r! = (2r+2-3)r! = 2(r+1)! - 3(r!)$.
Sum $= 2((n+1)! - 1!) - 3(n! - 1!) = 2(n+1)! - 3(n!) + 1$.
$S_n = n(n+1)! - [2(n+1)! - 3(n!) + 1] = (n-2)(n+1)! + 3(n!) - 1$.
Actually,the standard result for this series is $(n-1)(n+1)! + 1$.

(C) Let $S = \frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$.
The numerator is $\sum_{k=1}^{n} (2k)^2 = 4 \sum_{k=1}^{n} k^2 = 4 \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}$.
The denominator is $\sum_{k=1}^{n} (2k-1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) = 4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n = \frac{n(2n+1)(2n-1)}{3}$.
Thus,$S = \frac{2n(n+1)(2n+1)/3}{n(2n+1)(2n-1)/3} = \frac{2(n+1)}{2n-1}$.
We are given $S < 1.01$,so $\frac{2n+2}{2n-1} < \frac{101}{100}$.
$100(2n+2) < 101(2n-1) \Rightarrow 200n + 200 < 202n - 101$.
$301 < 2n \Rightarrow n > 150.5$.
Since $n$ must be an integer,the minimum value is $n = 151$.