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

(A) The general term of the series is $T_n = \frac{1}{(3n-1)(3n+2)}$.
We can write $T_n = \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)$.
For $n=16$,the last term is $\frac{1}{(3(16)-1)(3(16)+2)} = \frac{1}{47 \cdot 50}$.
The sum $S_{16} = \sum_{n=1}^{16} \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)$.
$S_{16} = \frac{1}{3} \left[ (\frac{1}{2} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{8}) + \ldots + (\frac{1}{47} - \frac{1}{50}) \right]$.
$S_{16} = \frac{1}{3} \left[ \frac{1}{2} - \frac{1}{50} \right] = \frac{1}{3} \left[ \frac{25-1}{50} \right] = \frac{1}{3} \left[ \frac{24}{50} \right] = \frac{8}{50} = \frac{4}{25}$.

(B) $S_n = 1^2 - 2^2 + 3^2 - 4^2 + \ldots + (-1)^{n-1} n^2$ \\ $S_{77} = (1^2 - 2^2) + (3^2 - 4^2) + \ldots + (75^2 - 76^2) + 77^2$ \\ Using the identity $a^2 - b^2 = (a-b)(a+b)$,each pair $(k^2 - (k+1)^2) = (k - k - 1)(k + k + 1) = -(2k+1)$ \\ $S_{77} = -(3 + 7 + 11 + \ldots + 151) + 77^2$ \\ The sum inside the bracket is an arithmetic progression with $n = 38$ terms,first term $a = 3$,and last term $l = 151$ \\ Sum $= \frac{38}{2}(3 + 151) = 19 \times 154 = 2926$ \\ $S_{77} = -2926 + 5929 = 3003$

(C) The given series is $S_n = 4^3 + 8^3 + 12^3 + \ldots + (4n)^3$.
This can be written as $S_n = \sum_{r=1}^{n} (4r)^3$.
$S_n = \sum_{r=1}^{n} 64r^3 = 64 \sum_{r=1}^{n} r^3$.
Using the formula for the sum of cubes of the first $n$ natural numbers,$\sum_{r=1}^{n} r^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4}$.
Substituting this into the expression for $S_n$:
$S_n = 64 \times \frac{n^2(n+1)^2}{4} = 16n^2(n+1)^2$.
Comparing this with the given form $k n^2(n+1)^2$,we get $k = 16$.

(D) Let $S_n = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}}$.
For any $k \geq 1$,we have $\sqrt{k} \leq \sqrt{n}$,which implies $\frac{1}{\sqrt{k}} \geq \frac{1}{\sqrt{n}}$.
Summing this inequality from $k=1$ to $n$:
$S_n = \sum_{k=1}^{n} \frac{1}{\sqrt{k}} \geq \sum_{k=1}^{n} \frac{1}{\sqrt{n}}$.
$S_n \geq n \times \frac{1}{\sqrt{n}} = \sqrt{n}$.
Thus,$S_n \geq \sqrt{n}$ for all $n \in N$.

(D) The $n$-th term of the series is $T_n = n(n+1)(n+2)$.
Expanding this,we get $T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n$.
The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^3 + 3k^2 + 2k)$.
Using standard summation formulas:
$\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$.
Substituting these:
$S_n = \frac{n^2(n+1)^2}{4} + 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$
$S_n = \frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + (2n+1) + 2 \right]$
$S_n = \frac{n(n+1)}{2} \left[ \frac{n^2+n+4n+2+4}{2} \right] = \frac{n(n+1)(n^2+5n+6)}{4}$
Since $n^2+5n+6 = (n+2)(n+3)$,we have $S_n = \frac{n(n+1)(n+2)(n+3)}{4}$.

(C) We need to evaluate the sum $S = \sum_{k=1}^n k(k+2)$.
Expanding the term inside the summation,we get $k^2 + 2k$.
Thus,$S = \sum_{k=1}^n (k^2 + 2k) = \sum_{k=1}^n k^2 + 2 \sum_{k=1}^n k$.
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 have:
$S = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$.
$S = \frac{n(n+1)(2n+1)}{6} + n(n+1)$.
Factoring out $\frac{n(n+1)}{6}$,we get:
$S = \frac{n(n+1)}{6} [ (2n+1) + 6 ]$.
$S = \frac{n(n+1)(2n+7)}{6}$.

(A) We know that the sum of the first $k$ cubes is $\left(\frac{k(k+1)}{2}\right)^2$ and the sum of the first $k$ odd numbers is $k^2$.
Substituting these into the expression:
$\sum_{k=1}^5 \frac{\left(\frac{k(k+1)}{2}\right)^2}{k^2} = \sum_{k=1}^5 \frac{k^2(k+1)^2}{4k^2} = \sum_{k=1}^5 \frac{(k+1)^2}{4}$
Expanding the sum for $k=1$ to $5$:
$= \frac{1}{4} [2^2 + 3^2 + 4^2 + 5^2 + 6^2] = \frac{1}{4} [4 + 9 + 16 + 25 + 36] = \frac{90}{4} = 22.5$

(D) The given series is $S_n = 2^3 + 4^3 + 6^3 + \ldots + (2n)^3$.
We can write this as $S_n = \sum_{k=1}^{n} (2k)^3$.
$S_n = \sum_{k=1}^{n} 8k^3 = 8 \sum_{k=1}^{n} k^3$.
Using the formula for the sum of cubes of the first $n$ natural numbers,$\sum_{k=1}^{n} k^3 = [\frac{n(n+1)}{2}]^2$.
So,$S_n = 8 \times [\frac{n(n+1)}{2}]^2 = 8 \times \frac{n^2(n+1)^2}{4} = 2n^2(n+1)^2$.
Comparing this with the given expression $h n^2(n+1)^2$,we get $h = 2$.

(A) Let $S_n = \sum_{k=1}^n k(k+1)(k+2) \ldots(k+r-1)$.
We use the property of falling factorials: $k(k+1) \ldots (k+r-1) = \frac{k(k+1) \ldots (k+r) - (k-1)k(k+1) \ldots (k+r-1)}{r+1}$.
Let $f(k) = k(k+1) \ldots (k+r-1)$. Then $f(k) = \frac{1}{r+1} [g(k) - g(k-1)]$,where $g(k) = k(k+1) \ldots (k+r)$.
Summing from $k=1$ to $n$ gives a telescoping sum:
$S_n = \frac{1}{r+1} \sum_{k=1}^n [g(k) - g(k-1)] = \frac{1}{r+1} [g(n) - g(0)]$.
Since $g(0) = 0 \cdot 1 \ldots r = 0$,we have $S_n = \frac{g(n)}{r+1} = \frac{n(n+1)(n+2) \ldots(n+r)}{r+1}$.

(C) We need to evaluate the sum $\sum_{n=1}^5 (n^3 + n^2 + n)$.
For $n=1$: $1(1^2+1+1) = 1(3) = 3$.
For $n=2$: $2(2^2+2+1) = 2(7) = 14$.
For $n=3$: $3(3^2+3+1) = 3(13) = 39$.
For $n=4$: $4(4^2+4+1) = 4(21) = 84$.
For $n=5$: $5(5^2+5+1) = 5(31) = 155$.
Summing these values: $3 + 14 + 39 + 84 + 155 = 295$.

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

(D) We know that the sum of binomial coefficients is $\sum_{r=0}^k \binom{k}{r} = 2^k$.
Substituting this into the given expression:
$\sum_{k=1}^{\infty} \frac{1}{3^k} \sum_{r=0}^k \binom{k}{r} = \sum_{k=1}^{\infty} \frac{1}{3^k} (2^k)$
$= \sum_{k=1}^{\infty} \left(\frac{2}{3}\right)^k$
This is an infinite geometric series with first term $a = 2/3$ and common ratio $r = 2/3$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
$S = \frac{2/3}{1 - 2/3} = \frac{2/3}{1/3} = 2$.

(C) The first $n$ natural numbers are $1, 2, 3, \ldots, n$.
Their squares are $1^2, 2^2, 3^2, \ldots, n^2$.
$\text{Mean} = \frac{\text{Sum of observations}}{\text{Total number of observations}}$.
$\text{Mean} = \frac{1^2+2^2+3^2+\ldots+n^2}{n}$.
Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$\text{Mean} = \frac{\frac{n(n+1)(2n+1)}{6}}{n} = \frac{(n+1)(2n+1)}{6}$.
Expanding the numerator: $\frac{2n^2+n+2n+1}{6} = \frac{2n^2+3n+1}{6}$.

(A) Given the recurrence relation $f(x)=f(x-2)+f(x-3)$ with initial values $f(0)=0, f(1)=1, f(2)=2$.
We calculate the subsequent values step by step:
$f(3)=f(1)+f(0)=1+0=1$
$f(4)=f(2)+f(1)=2+1=3$
$f(5)=f(3)+f(2)=1+2=3$
$f(6)=f(4)+f(3)=3+1=4$
$f(7)=f(5)+f(4)=3+3=6$
$f(8)=f(6)+f(5)=4+3=7$
$f(9)=f(7)+f(6)=6+4=10$
$f(10)=f(8)+f(7)=7+6=13$

(A) The $n$-th term of the series is $T_n = (2n-1)(2n+1)(2n+3) = (4n^2-1)(2n+3) = 8n^3 + 12n^2 - 2n - 3$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = 8 \sum k^3 + 12 \sum k^2 - 2 \sum k - 3 \sum 1$.
$S_n = 8 \left[ \frac{n(n+1)}{2} \right]^2 + 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 2 \left[ \frac{n(n+1)}{2} \right] - 3n$.
$S_n = 2n^2(n+1)^2 + 2n(n+1)(2n+1) - n(n+1) - 3n$.
$S_n = n(n+1) [2n(n+1) + 2(2n+1) - 1] - 3n$.
$S_n = n(n+1) [2n^2 + 2n + 4n + 2 - 1] - 3n = n(n+1)(2n^2 + 6n + 1) - 3n$.
Comparing with $n(n+1)f(n) - 3n$,we get $f(n) = 2n^2 + 6n + 1$.
Therefore,$f(1) = 2(1)^2 + 6(1) + 1 = 2 + 6 + 1 = 9$.

(C) We have,$1^4+2^4+3^4+\ldots+n^4 = f(n) \left(1^2+2^2+3^2+\ldots+n^2\right)$.
$f(n) = \frac{1^4+2^4+3^4+\ldots+n^4}{1^2+2^2+3^2+\ldots+n^2}$.
For $n=4$,we have $f(4) = \frac{1^4+2^4+3^4+4^4}{1^2+2^2+3^2+4^2}$.
$f(4) = \frac{1+16+81+256}{1+4+9+16}$.
$f(4) = \frac{354}{30}$.
Dividing both numerator and denominator by $6$,we get $f(4) = \frac{59}{5}$.

(A) The $n$-th term of the series is the sum of the first $n$ odd numbers,which is $T_n = n^2$.
We need to find the sum of the first $10$ terms: $S_{10} = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} n^2$.
The formula for the sum of the squares of the first $n$ natural numbers is $\frac{n(n+1)(2n+1)}{6}$.
Substituting $n=10$: $S_{10} = \frac{10(10+1)(2 \times 10 + 1)}{6} = \frac{10 \times 11 \times 21}{6}$.
$S_{10} = \frac{2310}{6} = 385$.

(D) The given series is $S_n = 1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \dots$ to $n$ terms.
For $n=2$,the sum is $S_2 = (1 \cdot 3 \cdot 5) + (3 \cdot 5 \cdot 7) = 15 + 105 = 120$.
According to the given formula,$S_n = n(n+1) f(n)$.
Substituting $n=2$:
$S_2 = 2(2+1) f(2) = 2(3) f(2) = 6 f(2)$.
Equating the two values:
$6 f(2) = 120$.
$f(2) = \frac{120}{6} = 20$.

(D) The general term of the series is $T_r = r(n-r+1) = nr - r^2 + r$.
Summing from $r=1$ to $n$:
$S_n = \sum_{r=1}^n (nr - r^2 + r) = (n+1) \sum_{r=1}^n r - \sum_{r=1}^n r^2$.
Using standard summation formulas:
$S_n = (n+1) \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$.
Factoring out $\frac{n(n+1)}{2}$:
$S_n = \frac{n(n+1)}{2} [ (n+1) - \frac{2n+1}{3} ] = \frac{n(n+1)}{2} [ \frac{3n+3-2n-1}{3} ] = \frac{n(n+1)(n+2)}{6}$.
Comparing this with $\alpha n(n+1)(n+2)$,we get $\alpha = \frac{1}{6}$.

(A) Given the recurrence relation $a_n = a_{n-1} + a_{n-2}$ with $a_0 = 0$ and $a_1 = 1$.
This is the characteristic equation $r^2 - r - 1 = 0$.
The roots are $r = \frac{1 \pm \sqrt{5}}{2}$.
Let $a_n = A\left(\frac{1+\sqrt{5}}{2}\right)^n + B\left(\frac{1-\sqrt{5}}{2}\right)^n$.
Using $a_0 = 0$,we get $A + B = 0$,so $B = -A$.
Using $a_1 = 1$,we get $A\left(\frac{1+\sqrt{5}}{2}\right) - A\left(\frac{1-\sqrt{5}}{2}\right) = 1$.
$A\left(\frac{1+\sqrt{5}-1+\sqrt{5}}{2}\right) = 1$ $\Rightarrow A\sqrt{5} = 1$ $\Rightarrow A = \frac{1}{\sqrt{5}}$.
Thus,$B = -\frac{1}{\sqrt{5}}$.
Substituting these values,$a_n = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n$.

(C) The series is $1^2, 2(2^2), 3^2, 2(4^2), 5^2, 2(6^2), \ldots$
For even $n$,let $n = 2m$. The sum $S_{2m}$ consists of $m$ odd-indexed terms and $m$ even-indexed terms.
$S_{2m} = (1^2 + 3^2 + 5^2 + \ldots + (2m-1)^2) + 2(2^2 + 4^2 + 6^2 + \ldots + (2m)^2)$
$= \sum_{k=1}^{m} (2k-1)^2 + 2 \sum_{k=1}^{m} (2k)^2$
$= \sum_{k=1}^{m} (4k^2 - 4k + 1) + 8 \sum_{k=1}^{m} k^2$
$= 12 \sum_{k=1}^{m} k^2 - 4 \sum_{k=1}^{m} k + \sum_{k=1}^{m} 1$
$= 12 \frac{m(m+1)(2m+1)}{6} - 4 \frac{m(m+1)}{2} + m$
$= 2m(m+1)(2m+1) - 2m(m+1) + m$
$= 2m(m+1)(2m+1-1) + m$
$= 2m(m+1)(2m) + m = 4m^2(m+1) + m = m(4m^2 + 4m + 1) = m(2m+1)^2$
Since $n = 2m$,we have $m = n/2$.
Substituting $m = n/2$ into the expression:
$S_n = \frac{n}{2}(2(\frac{n}{2}) + 1)^2 = \frac{n}{2}(n+1)^2$.

(B) Consider the sum: $\sum_{k=1}^n k(k+2) = \sum_{k=1}^n (k^2 + 2k)$.
Using the standard summation formulas $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$:
$\sum_{k=1}^n k^2 + 2 \sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$.
Factor out $\frac{n(n+1)}{6}$:
$= \frac{n(n+1)}{6} [ (2n+1) + 6 ]$.
$= \frac{n(n+1)(2n+7)}{6}$.

(C) Given,$S_n = \sum_{k=1}^{n} k^3$ and $T_n = \sum_{k=1}^{n} k$.
We know that the sum of the first $n$ natural numbers is $T_n = \frac{n(n+1)}{2}$.
The sum of the cubes of the first $n$ natural numbers is $S_n = \left[\frac{n(n+1)}{2}\right]^2$.
Substituting $T_n$ into the expression for $S_n$,we get $S_n = (T_n)^2$.
Therefore,the correct relation is $S_n = T_n^2$.

(D) The given series is $S_n = 2^3 + 4^3 + 6^3 + \ldots + (2n)^3$.
We can write this as $S_n = \sum_{k=1}^{n} (2k)^3$.
$S_n = \sum_{k=1}^{n} 8k^3 = 8 \sum_{k=1}^{n} k^3$.
Using the formula for the sum of cubes of the first $n$ natural numbers,$\sum_{k=1}^{n} k^3 = [\frac{n(n+1)}{2}]^2$.
Thus,$S_n = 8 \times [\frac{n(n+1)}{2}]^2 = 8 \times \frac{n^2(n+1)^2}{4} = 2n^2(n+1)^2$.
Comparing this with the given expression $h n^2(n+1)^2$,we get $h = 2$.

(A) Observe the pattern of the sums:
$S_1 = 1^2 = 1$
$S_2 = 3^2 = 9$
$S_3 = 6^2 = 36$
$S_4 = 10^2 = 100$
$S_5 = 15^2 = 225$
The bases of the squares are $1, 3, 6, 10, 15, \ldots$.
These are the triangular numbers,given by the formula $T_n = \frac{n(n+1)}{2}$.
For $n = 10$,the base $k$ is the $10^{th}$ triangular number:
$k = T_{10} = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = 55$.
Thus,$S_{10} = 55^2$,so $k = 55$.

(B) The given sum is $S = 1^2 - 2^2 + 3^2 - 4^2 + \dots - (2n)^2 + (2n+1)^2$.
We can group the terms as:
$S = (1^2 - 2^2) + (3^2 - 4^2) + \dots + ((2n-1)^2 - (2n)^2) + (2n+1)^2$.
Using the identity $a^2 - b^2 = (a-b)(a+b)$,each pair becomes:
$(1-2)(1+2) + (3-4)(3+4) + \dots + ((2n-1)-2n)((2n-1)+2n) + (2n+1)^2$.
$S = -1(3) - 1(7) - 1(11) - \dots - 1(4n-1) + (2n+1)^2$.
$S = -(3 + 7 + 11 + \dots + (4n-1)) + (2n+1)^2$.
The sum inside the bracket is an arithmetic progression with $n$ terms,first term $a=3$,and last term $l=4n-1$.
Sum $= \frac{n}{2}(3 + 4n - 1) = \frac{n}{2}(4n+2) = n(2n+1)$.
Thus,$S = -n(2n+1) + (2n+1)^2$.
$S = (2n+1)(-n + 2n + 1) = (2n+1)(n+1)$.

(B) Given,$a_0=1$ and $a_{n+1}=3n^2+n+a_n$.
We can find the first few terms:
$a_1 = 3(0)^2 + 0 + a_0 = 0 + 0 + 1 = 1$.
$a_2 = 3(1)^2 + 1 + a_1 = 3 + 1 + 1 = 5$.
$a_3 = 3(2)^2 + 2 + a_2 = 12 + 2 + 5 = 19$.
Now,check the options:
For $n=0$: $a_0 = 0^3 - 0^2 + 1 = 1$ (Matches).
For $n=1$: $a_1 = 1^3 - 1^2 + 1 = 1$ (Matches).
For $n=2$: $a_2 = 2^3 - 2^2 + 1 = 8 - 4 + 1 = 5$ (Matches).
For $n=3$: $a_3 = 3^3 - 3^2 + 1 = 27 - 9 + 1 = 19$ (Matches).
Thus,$a_n = n^3 - n^2 + 1$ is the correct expression.

(D) We are given the expression $\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$.
Using the identity $\sum_{r=0}^k \binom{k}{r} = 2^k$,we can simplify the inner summation:
$\sum_{k=1}^{\infty} \frac{1}{3^k} \left( \sum_{r=0}^k \binom{k}{r} \right) = \sum_{k=1}^{\infty} \frac{2^k}{3^k} = \sum_{k=1}^{\infty} \left( \frac{2}{3} \right)^k$.
This is an infinite geometric series with the first term $a = \frac{2}{3}$ and common ratio $r = \frac{2}{3}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
Substituting the values,we get $S = \frac{2/3}{1 - 2/3} = \frac{2/3}{1/3} = 2$.

(B) The given series is $S = 1 + \frac{\cos \theta}{2} + \frac{\cos 2 \theta}{4} + \frac{\cos 3 \theta}{8} + \ldots$
This is an infinite series of the form $\sum_{n=0}^{\infty} \frac{\cos(n\theta)}{2^n}$.
Using the formula $\sum_{n=0}^{\infty} r^n \cos(n\theta) = \frac{1-r \cos \theta}{1-2r \cos \theta + r^2}$ with $r = \frac{1}{2}$:
$S = \frac{1-\frac{1}{2} \cos \theta}{1-2(\frac{1}{2}) \cos \theta + (\frac{1}{2})^2} = \frac{1-\frac{1}{2} \cos \theta}{1-\cos \theta + \frac{1}{4}} = \frac{\frac{2-\cos \theta}{2}}{\frac{5-4 \cos \theta}{4}} = \frac{2(2-\cos \theta)}{5-4 \cos \theta} = \frac{4-2 \cos \theta}{5-4 \cos \theta}$.
Comparing this with $\frac{a-2 \cos \theta}{5+b \cos \theta}$,we get $a = 4$ and $b = -4$.
Therefore,$(a-b)^2 = (4 - (-4))^2 = (8)^2 = 64$.

(C) Given $f(1)=3$ and $f(n+1)-f(n)=3(4^n-1)$.
We can write the recurrence as a telescoping sum:
$f(n) = f(1) + \sum_{k=1}^{n-1} (f(k+1)-f(k))$
$f(n) = 3 + \sum_{k=1}^{n-1} 3(4^k-1)$
$f(n) = 3 + 3 \left( \sum_{k=1}^{n-1} 4^k - \sum_{k=1}^{n-1} 1 \right)$
Using the geometric series sum formula $\sum_{k=1}^{n-1} 4^k = \frac{4(4^{n-1}-1)}{4-1} = \frac{4^n-4}{3}$:
$f(n) = 3 + 3 \left( \frac{4^n-4}{3} - (n-1) \right)$
$f(n) = 3 + (4^n-4) - 3(n-1)$
$f(n) = 3 + 4^n - 4 - 3n + 3$
$f(n) = 4^n - 3n + 2$

(C) The given series is $S = \sum_{n=1}^{\infty} (1+a+a^2+\dots+a^{n-1})x^{n-1}$.
Using the sum of a geometric progression,$1+a+a^2+\dots+a^{n-1} = \frac{1-a^n}{1-a}$.
So,$S = \sum_{n=1}^{\infty} \frac{1-a^n}{1-a} x^{n-1} = \frac{1}{1-a} \sum_{n=1}^{\infty} (x^{n-1} - a^n x^{n-1})$.
$S = \frac{1}{1-a} \left( \sum_{n=1}^{\infty} x^{n-1} - a \sum_{n=1}^{\infty} (ax)^{n-1} \right)$.
Using the sum of an infinite geometric series $\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$ for $|r| < 1$:
$S = \frac{1}{1-a} \left( \frac{1}{1-x} - \frac{a}{1-ax} \right)$.
$S = \frac{1}{1-a} \left( \frac{(1-ax) - a(1-x)}{(1-x)(1-ax)} \right) = \frac{1}{1-a} \left( \frac{1-ax-a+ax}{(1-x)(1-ax)} \right)$.
$S = \frac{1}{1-a} \left( \frac{1-a}{(1-x)(1-ax)} \right) = \frac{1}{(1-x)(1-ax)}$.

(A) The $n^{th}$ term of the series is $T_{n} = (2n-1)^{3}$.
Expanding this,we get $T_{n} = 8n^{3} - 12n^{2} + 6n - 1$.
The sum of $n$ terms is $S_{n} = \sum_{k=1}^{n} T_{k} = \sum_{k=1}^{n} (8k^{3} - 12k^{2} + 6k - 1)$.
Using standard summation formulas:
$S_{n} = 8 \left[ \frac{n(n+1)}{2} \right]^{2} - 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 6 \left[ \frac{n(n+1)}{2} \right] - n$.
$S_{n} = 2n^{2}(n+1)^{2} - 2n(n+1)(2n+1) + 3n(n+1) - n$.
$S_{n} = 2n^{2}(n^{2}+2n+1) - 2n(2n^{2}+3n+1) + 3n^{2} + 3n - n$.
$S_{n} = 2n^{4} + 4n^{3} + 2n^{2} - 4n^{3} - 6n^{2} - 2n + 3n^{2} + 2n$.
$S_{n} = 2n^{4} - n^{2} = n^{2}(2n^{2}-1)$.

(B) We have $a_n = (\frac{n(n+1)(2n+1)}{6})^n$ and $b_n = n^n(n!)$.
For $n=1$,$a_1 = (1^2)^1 = 1$ and $b_1 = 1^1(1!) = 1$,so $a_1 = b_1$.
For $n=2$,$a_2 = (1^2+2^2)^2 = 5^2 = 25$ and $b_2 = 2^2(2!) = 4 \times 2 = 8$. Thus $a_2 > b_2$.
For $n=3$,$a_3 = (1^2+2^2+3^2)^3 = 14^3 = 2744$ and $b_3 = 3^3(3!) = 27 \times 6 = 162$. Thus $a_3 > b_3$.
By induction or comparing terms,it can be shown that $a_n > b_n$ for all $n \geq 2$.

(A) Let $a = \frac{4}{7}$ and $b = \frac{1}{3}$.
Each term is of the form $\sum_{k=0}^{n} a^k b^{n-k} = \frac{a^{n+1} - b^{n+1}}{a - b}$.
The series is $\sum_{n=1}^{\infty} \frac{a^{n+1} - b^{n+1}}{a - b} = \frac{1}{a - b} \left[ \sum_{n=1}^{\infty} a^{n+1} - \sum_{n=1}^{\infty} b^{n+1} \right]$.
Here $a - b = \frac{4}{7} - \frac{1}{3} = \frac{12 - 7}{21} = \frac{5}{21}$.
Sum $= \frac{21}{5} \left[ \frac{a^2}{1 - a} - \frac{b^2}{1 - b} \right] = \frac{21}{5} \left[ \frac{(4/7)^2}{1 - 4/7} - \frac{(1/3)^2}{1 - 1/3} \right]$.
Sum $= \frac{21}{5} \left[ \frac{16/49}{3/7} - \frac{1/9}{2/3} \right] = \frac{21}{5} \left[ \frac{16}{21} - \frac{1}{6} \right]$.
Sum $= \frac{21}{5} \left[ \frac{32 - 7}{42} \right] = \frac{21}{5} \times \frac{25}{42} = \frac{1}{1} \times \frac{5}{2} = \frac{5}{2}$.

(C) The $n$-th term of the series is given by $T_n = \frac{1}{n} \sum_{k=1}^n k^2$.
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,$T_n = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6} = \frac{2n^2+3n+1}{6} = \frac{n^2}{3} + \frac{n}{2} + \frac{1}{6}$.
To find the sum of the first $10$ terms,we calculate $S_{10} = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} (\frac{n^2}{3} + \frac{n}{2} + \frac{1}{6})$.
$S_{10} = \frac{1}{3} \sum_{n=1}^{10} n^2 + \frac{1}{2} \sum_{n=1}^{10} n + \sum_{n=1}^{10} \frac{1}{6}$.
$S_{10} = \frac{1}{3} \left( \frac{10 \cdot 11 \cdot 21}{6} \right) + \frac{1}{2} \left( \frac{10 \cdot 11}{2} \right) + \frac{10}{6}$.
$S_{10} = \frac{1}{3} (385) + \frac{55}{2} + \frac{5}{3} = \frac{385+5}{3} + 27.5 = \frac{390}{3} + 27.5 = 130 + 27.5 = 157.5$.
Since $157.5 = \frac{315}{2}$,the correct option is $C$.

(C) First,find the values of $\alpha$ and $\beta$ using the sum of an infinite geometric progression formula $S = \frac{a}{1-r}$.
For $\alpha$: $a = 1/4$,$r = 1/2$,so $\alpha = \frac{1/4}{1-1/2} = 1/2$.
For $\beta$: $a = 1/3$,$r = 1/3$,so $\beta = \frac{1/3}{1-1/3} = 1/2$.
Now,substitute these values into the expression: $(0.2)^{\log_{\sqrt{5}}(1/2)} + (0.04)^{\log_{5}(1/2)}$.
Note that $0.2 = 5^{-1}$ and $0.04 = 5^{-2}$.
For the first term: $\log_{\sqrt{5}}(1/2) = \frac{\log_5(1/2)}{\log_5(5^{1/2})} = \frac{-\log_5(2)}{1/2} = -2 \log_5(2)$.
Thus,$(5^{-1})^{-2 \log_5(2)} = 5^{2 \log_5(2)} = 5^{\log_5(2^2)} = 2^2 = 4$.
For the second term: $(5^{-2})^{\log_5(1/2)} = (5^{\log_5(1/2)})^{-2} = (1/2)^{-2} = 2^2 = 4$.
Summing the results: $4 + 4 = 8$.

(B) The $n$-th term is $T_n = \frac{\sum_{i=1}^n i^3}{\sum_{i=1}^n (2i-1)}$.
We know that $\sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$ and $\sum_{i=1}^n (2i-1) = n^2$.
Thus,$T_n = \frac{(\frac{n(n+1)}{2})^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4} = \frac{n^2+2n+1}{4}$.
The sum of $8$ terms is $S_8 = \sum_{n=1}^8 \frac{n^2+2n+1}{4} = \frac{1}{4} [ \sum_{n=1}^8 n^2 + 2\sum_{n=1}^8 n + \sum_{n=1}^8 1 ]$.
Using the formulas $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$:
$S_8 = \frac{1}{4} [ \frac{8(9)(17)}{6} + 2(\frac{8(9)}{2}) + 8 ] = \frac{1}{4} [ 204 + 72 + 8 ] = \frac{284}{4} = 71$.

(B) Let the series be $S = 1^3 - 2^3 + 3^3 - 4^3 + \dots + 15^3$.
We can group the terms as $S = (1^3 + 3^3 + \dots + 15^3) - (2^3 + 4^3 + \dots + 14^3)$.
The sum of odd cubes is $\sum_{k=1}^8 (2k-1)^3 = \sum_{k=1}^8 (8k^3 - 12k^2 + 6k - 1)$.
Using standard summation formulas:
$\sum_{k=1}^8 k^3 = [8(9)/2]^2 = 36^2 = 1296$.
$\sum_{k=1}^8 k^2 = 8(9)(17)/6 = 204$.
$\sum_{k=1}^8 k = 8(9)/2 = 36$.
Sum of odd cubes $= 8(1296) - 12(204) + 6(36) - 8 = 10368 - 2448 + 216 - 8 = 8128$.
The sum of even cubes is $\sum_{k=1}^7 (2k)^3 = 8 \sum_{k=1}^7 k^3 = 8 [7(8)/2]^2 = 8(28^2) = 8(784) = 6272$.
Therefore,$S = 8128 - 6272 = 1856$.

(A) Given $S_n = \sum_{k=1}^{n} a_k = 6n^3$.
We know that $a_n = S_n - S_{n-1}$ for $n > 1$.
$a_n = 6n^3 - 6(n-1)^3 = 6(n^3 - (n^3 - 3n^2 + 3n - 1)) = 6(3n^2 - 3n + 1) = 18n^2 - 18n + 6$.
Now,calculate $a_{k+1} - a_k$:
$a_{k+1} - a_k = [18(k+1)^2 - 18(k+1) + 6] - [18k^2 - 18k + 6]$
$= 18(k^2 + 2k + 1 - k^2) - 18(k + 1 - k) = 18(2k + 1) - 18 = 36k$.
Substitute this into the given expression:
$\sum_{k=1}^{6} \left(\frac{36k}{36}\right)^2 = \sum_{k=1}^{6} k^2$.
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.
For $n=6$,$\sum_{k=1}^{6} k^2 = \frac{6(6+1)(2 \cdot 6 + 1)}{6} = 7 \cdot 13 = 91$.