Arithmetic geometric progression,Method of difference Questions in English

(B) The given series is an arithmetico-geometric series: $S = 1 + \frac{3}{2} + \frac{5}{2^2} + \frac{7}{2^3} + \dots \infty$.
Here,the arithmetic part is $1, 3, 5, 7, \dots$ with first term $a = 1$ and common difference $d = 2$.
The geometric part is $1, \frac{1}{2}, \frac{1}{2^2}, \frac{1}{2^3}, \dots$ with common ratio $r = \frac{1}{2}$.
The sum of an infinite arithmetico-geometric series is given by $S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$.
Substituting the values: $S_{\infty} = \frac{1}{1 - \frac{1}{2}} + \frac{2 \times \frac{1}{2}}{(1 - \frac{1}{2})^2}$.
$S_{\infty} = \frac{1}{\frac{1}{2}} + \frac{1}{(\frac{1}{2})^2} = 2 + \frac{1}{\frac{1}{4}} = 2 + 4 = 6$.
Wait,let us re-evaluate: $S = 1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \dots$
$\frac{1}{2}S = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \dots$
Subtracting: $S - \frac{1}{2}S = 1 + (\frac{3}{2} - \frac{1}{2}) + (\frac{5}{4} - \frac{3}{4}) + (\frac{7}{8} - \frac{5}{8}) + \dots$
$\frac{1}{2}S = 1 + 1 + \frac{2}{4} + \frac{2}{8} + \dots = 1 + (1 + \frac{1}{2} + \frac{1}{4} + \dots) = 1 + \frac{1}{1 - \frac{1}{2}} = 1 + 2 = 3$.
Therefore,$S = 3 \times 2 = 6$.

(D) Let the sum to infinity of the arithmetico-geometric series be $S = 1 + 4 \cdot \frac{1}{5} + 7 \cdot \frac{1}{5^2} + 10 \cdot \frac{1}{5^3} + \dots$
$\frac{1}{5}S = \frac{1}{5} + 4 \cdot \frac{1}{5^2} + 7 \cdot \frac{1}{5^3} + \dots$
Subtracting the two equations:
$(1 - \frac{1}{5})S = 1 + 3 \cdot \frac{1}{5} + 3 \cdot \frac{1}{5^2} + 3 \cdot \frac{1}{5^3} + \dots$
$\frac{4}{5}S = 1 + 3 \left( \frac{1}{5} + \frac{1}{5^2} + \dots \right)$
Using the sum of an infinite geometric series formula $S_{\infty} = \frac{a}{1-r}$ where $a = \frac{1}{5}$ and $r = \frac{1}{5}$:
$\frac{4}{5}S = 1 + 3 \left( \frac{1/5}{1 - 1/5} \right) = 1 + 3 \left( \frac{1/5}{4/5} \right) = 1 + 3 \left( \frac{1}{4} \right) = 1 + \frac{3}{4} = \frac{7}{4}$
$S = \frac{7}{4} \times \frac{5}{4} = \frac{35}{16}$.

(A) Let $S_n$ be the sum of the given series to $n$ terms:
$S_n = 1 + 2x + 3x^2 + 4x^3 + \dots + nx^{n - 1}$ $(i)$
Multiply by $x$:
$xS_n = x + 2x^2 + 3x^3 + \dots + (n - 1)x^{n - 1} + nx^n$ $(ii)$
Subtracting $(ii)$ from $(i)$:
$(1 - x)S_n = 1 + x + x^2 + x^3 + \dots + x^{n - 1} - nx^n$
Using the sum formula for a geometric progression with $n$ terms:
$(1 - x)S_n = \frac{1 - x^n}{1 - x} - nx^n$
$(1 - x)S_n = \frac{1 - x^n - nx^n(1 - x)}{1 - x}$
$(1 - x)S_n = \frac{1 - x^n - nx^n + nx^{n + 1}}{1 - x}$
$(1 - x)S_n = \frac{1 - (n + 1)x^n + nx^{n + 1}}{1 - x}$
Therefore,$S_n = \frac{1 - (n + 1)x^n + nx^{n + 1}}{(1 - x)^2}$.

(A) Let the sum be $S_n = 1 + \frac{2}{5} + \frac{3}{5^2} + \frac{4}{5^3} + \dots + \frac{n}{5^{n-1}}$.
Multiply by $\frac{1}{5}$:
$\frac{1}{5}S_n = \frac{1}{5} + \frac{2}{5^2} + \frac{3}{5^3} + \dots + \frac{n}{5^n}$.
Subtracting the two equations:
$(1 - \frac{1}{5})S_n = 1 + (\frac{2}{5} - \frac{1}{5}) + (\frac{3}{5^2} - \frac{2}{5^2}) + \dots + (\frac{n}{5^{n-1}} - \frac{n-1}{5^{n-1}}) - \frac{n}{5^n}$.
$\frac{4}{5}S_n = 1 + \frac{1}{5} + \frac{1}{5^2} + \dots + \frac{1}{5^{n-1}} - \frac{n}{5^n}$.
The sum of the geometric progression is $\frac{1(1 - (1/5)^n)}{1 - 1/5} = \frac{1 - 1/5^n}{4/5} = \frac{5}{4}(1 - \frac{1}{5^n})$.
$\frac{4}{5}S_n = \frac{5}{4}(1 - \frac{1}{5^n}) - \frac{n}{5^n}$.
$S_n = \frac{25}{16}(1 - \frac{1}{5^n}) - \frac{5n}{4 \times 5^n} = \frac{25}{16} - \frac{25}{16 \times 5^n} - \frac{20n}{16 \times 5^n}$.
$S_n = \frac{25}{16} - \frac{25 + 20n}{16 \times 5^n} = \frac{25}{16} - \frac{5(5 + 4n)}{16 \times 5 \times 5^{n-1}} = \frac{25}{16} - \frac{4n + 5}{16 \times 5^{n-1}}$.

(C) The given series is an Arithmetico-Geometric progression $(A.G.P.)$.
The numerator terms are $1, 4, 7, 10, \dots$,which form an Arithmetic Progression $(A.P.)$ with first term $a = 1$ and common difference $d = 3$.
The $n^{th}$ term of this $A.P.$ is $T_n = a + (n - 1)d = 1 + (n - 1)3 = 3n - 2$.
The denominator terms are $1, 5, 5^2, 5^3, \dots$,which form a Geometric Progression $(G.P.)$ with first term $a = 1$ and common ratio $r = 5$.
The $n^{th}$ term of this $G.P.$ is $G_n = ar^{n-1} = 1 \cdot 5^{n-1} = 5^{n-1}$.
Combining these,the $n^{th}$ term of the series is $\frac{3n - 2}{5^{n - 1}}$.

(C) Let $t = 1 + \frac{1}{n}$. The given sum is $S = \sum_{k=1}^n k t^{k-1} = 1 + 2t + 3t^2 + \dots + nt^{n-1}$.
This is an arithmetico-geometric series.
For an infinite series,$\sum_{k=1}^{\infty} k t^{k-1} = (1-t)^{-2}$.
Substituting $t = 1 + \frac{1}{n}$,we get $1-t = 1 - (1 + \frac{1}{n}) = -\frac{1}{n}$.
Thus,the sum is $(-\frac{1}{n})^{-2} = n^2$.

(A) Given the series $S_n = 1 + 2x + 3x^2 + 4x^3 + \dots + nx^{n-1} \dots (1)$
Multiply by $x$: $xS_n = x + 2x^2 + 3x^3 + \dots + (n-1)x^{n-1} + nx^n \dots (2)$
Subtracting $(2)$ from $(1)$:
$(1 - x)S_n = 1 + (2x - x) + (3x^2 - 2x^2) + \dots + (nx^{n-1} - (n-1)x^{n-1}) - nx^n$
$(1 - x)S_n = 1 + x + x^2 + \dots + x^{n-1} - nx^n$
The sum of the geometric series $1 + x + x^2 + \dots + x^{n-1}$ is $\frac{1 - x^n}{1 - x}$.
So,$(1 - x)S_n = \frac{1 - x^n}{1 - x} - nx^n$
$(1 - x)S_n = \frac{1 - x^n - nx^n(1 - x)}{1 - x}$
$S_n = \frac{1 - x^n - nx^n + nx^{n+1}}{(1 - x)^2}$
$S_n = \frac{1 - (n + 1)x^n + nx^{n+1}}{(1 - x)^2}$

(C) The $r^{th}$ term is given by $T_r = (2r + 1)2^{-r} = \frac{2r + 1}{2^r}$.
The sum of infinite terms is $S_{\infty} = \sum_{r=1}^{\infty} \frac{2r + 1}{2^r} = \sum_{r=1}^{\infty} \frac{2r}{2^r} + \sum_{r=1}^{\infty} \frac{1}{2^r}$.
Let $S = \sum_{r=1}^{\infty} \frac{2r + 1}{2^r} = \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \frac{9}{16} + \dots$
This is an arithmetico-geometric series. Multiplying by $\frac{1}{2}$:
$\frac{1}{2}S = \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + \dots$
Subtracting the two equations:
$S - \frac{1}{2}S = \frac{3}{2} + (\frac{5}{4} - \frac{3}{4}) + (\frac{7}{8} - \frac{5}{8}) + \dots$
$\frac{1}{2}S = \frac{3}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + \dots$
$\frac{1}{2}S = \frac{3}{2} + 2(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots)$
The term in the bracket is a geometric series with $a = \frac{1}{4}$ and $r = \frac{1}{2}$,so the sum is $\frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.
$\frac{1}{2}S = \frac{3}{2} + 2(\frac{1}{2}) = \frac{3}{2} + 1 = \frac{5}{2}$.
Therefore,$S = 5$.

(A) Let the sum of the infinite arithmetic-geometric series be $S = 1 + 4 \cdot \frac{1}{5} + 7 \cdot \frac{1}{5^2} + 10 \cdot \frac{1}{5^3} + \dots$
Multiply by $\frac{1}{5}$:
$\frac{1}{5}S = \frac{1}{5} + 4 \cdot \frac{1}{5^2} + 7 \cdot \frac{1}{5^3} + \dots$
Subtracting the two equations:
$S - \frac{1}{5}S = 1 + (4-1) \cdot \frac{1}{5} + (7-4) \cdot \frac{1}{5^2} + (10-7) \cdot \frac{1}{5^3} + \dots$
$\frac{4}{5}S = 1 + 3 \cdot \frac{1}{5} + 3 \cdot \frac{1}{5^2} + 3 \cdot \frac{1}{5^3} + \dots$
$\frac{4}{5}S = 1 + 3 \left( \frac{1}{5} + \frac{1}{5^2} + \frac{1}{5^3} + \dots \right)$
The term in the bracket is an infinite geometric series with $a = \frac{1}{5}$ and $r = \frac{1}{5}$:
Sum $= \frac{a}{1-r} = \frac{1/5}{1 - 1/5} = \frac{1/5}{4/5} = \frac{1}{4}$
So,$\frac{4}{5}S = 1 + 3 \left( \frac{1}{4} \right) = 1 + \frac{3}{4} = \frac{7}{4}$
$S = \frac{7}{4} \times \frac{5}{4} = \frac{35}{16}$

(C) Let $S = 1 + 2 \cdot 2 + 3 \cdot 2^2 + 4 \cdot 2^3 + \dots + 100 \cdot 2^{99} \quad \dots(1)$
Multiply by $2$:
$2S = 1 \cdot 2^2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + 99 \cdot 2^{99} + 100 \cdot 2^{100} \quad \dots(2)$
Subtract $(1)$ from $(2)$:
$2S - S = 100 \cdot 2^{100} - (1 + 2 + 2^2 + 2^3 + \dots + 2^{99})$
$S = 100 \cdot 2^{100} - \frac{1(2^{100} - 1)}{2 - 1}$
$S = 100 \cdot 2^{100} - 2^{100} + 1$
$S = 99 \cdot 2^{100} + 1$

(D) Let the series be $S_n = 12 + 16 + 24 + 40 + \dots + T_n$.
The differences between consecutive terms are $4, 8, 16, \dots$,which form a geometric progression.
The $n$-th term $T_n$ can be found using the method of differences:
$T_n = T_1 + \sum_{k=1}^{n-1} (T_{k+1} - T_k) = 12 + \sum_{k=1}^{n-1} 2^{k+1} = 12 + (4 + 8 + 16 + \dots + 2^n)$.
Using the sum of a geometric series formula,$T_n = 12 + \frac{4(2^{n-1} - 1)}{2 - 1} = 12 + 2^{n+1} - 4 = 2^{n+1} + 8$.
Now,the sum of $n$ terms $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (2^{k+1} + 8)$.
$S_n = (2^2 + 2^3 + \dots + 2^{n+1}) + (8 + 8 + \dots + 8)$.
$S_n = \frac{4(2^n - 1)}{2 - 1} + 8n = 4(2^n - 1) + 8n$.

(A) Let $S = 10^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \dots + 10(11)^9 = k(10)^9$.
Dividing both sides by $10^9$,we get:
$k = 1 + 2\left(\frac{11}{10}\right) + 3\left(\frac{11}{10}\right)^2 + \dots + 10\left(\frac{11}{10}\right)^9$ ......$(i)$
Let $x = \frac{11}{10}$. Then $k = 1 + 2x + 3x^2 + \dots + 10x^9$.
Multiply by $x$:
$xk = x + 2x^2 + 3x^3 + \dots + 9x^9 + 10x^{10}$ ......$(ii)$
Subtracting $(ii)$ from $(i)$:
$k(1-x) = 1 + x + x^2 + \dots + x^9 - 10x^{10}$
$k(1-x) = \frac{1(x^{10}-1)}{x-1} - 10x^{10}$
Since $x = \frac{11}{10}$,$1-x = -\frac{1}{10}$ and $x-1 = \frac{1}{10}$.
$k(-\frac{1}{10}) = \frac{(\frac{11}{10})^{10}-1}{\frac{1}{10}} - 10(\frac{11}{10})^{10}$
$k(-\frac{1}{10}) = 10((\frac{11}{10})^{10}-1) - 10(\frac{11}{10})^{10}$
$k(-\frac{1}{10}) = 10(\frac{11}{10})^{10} - 10 - 10(\frac{11}{10})^{10} = -10$
$k = 100$.

(D) The given series is of the form $S = 1 + 2x + 3x^2 + 4x^3 + \dots \infty$,where $x = (1 - \cos \theta)$.
This is an arithmetico-geometric series,and its sum is given by $S = (1 - x)^{-2}$ for $|x| < 1$.
Substituting $x = 1 - \cos \theta$,we get $S = (1 - (1 - \cos \theta))^{-2} = (\cos \theta)^{-2} = \sec^2 \theta$.
Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$,we have $S = 1 + \tan^2 \theta$.
Given $\tan \theta = \sqrt{\frac{3}{2}}$,then $\tan^2 \theta = \frac{3}{2}$.
Therefore,$S = 1 + \frac{3}{2} = \frac{5}{2}$.

(B) The given series is an arithmetico-geometric series: $f(x) = \sum_{n=1}^{\infty} n \sin^n x$.
For $|\sin x| < 1$,the sum is given by $f(x) = \frac{\sin x}{(1 - \sin x)^2}$.
Setting $f(x) = 2$,we have $\frac{\sin x}{(1 - \sin x)^2} = 2$.
$\sin x = 2(1 - 2\sin x + \sin^2 x) = 2 - 4\sin x + 2\sin^2 x$.
$2\sin^2 x - 5\sin x + 2 = 0$.
$(2\sin x - 1)(\sin x - 2) = 0$.
Since $\sin x = 2$ is impossible,we have $\sin x = \frac{1}{2}$.
In the interval $x \in [-\pi, \pi]$,$\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.
Both values are in the domain $[-\pi, \pi] - \{\pm \frac{\pi}{2}\}$.
Thus,there are $2$ solutions.

(D) Since $x, y, z$ are in $G.P.$,we have $y^2 = xz$.
Taking the natural logarithm on both sides,$2 \ln y = \ln x + \ln z$,which implies $\ln x, \ln y, \ln z$ are in $A.P.$
Let $a = \ln x, b = \ln y, c = \ln z$. Then $a, b, c$ are in $A.P.$
We are given the terms $T_1 = \frac{1+a}{2}, T_2 = \frac{1+b}{4}, T_3 = \frac{1+c}{8}$.
Since $a, b, c$ are in $A.P.$,let $a = A-d, b = A, c = A+d$.
Then $T_1 = \frac{1+A-d}{2}, T_2 = \frac{1+A}{4}, T_3 = \frac{1+A+d}{8}$.
Checking for $G.P.$: $T_2^2 = \frac{(1+A)^2}{16}$ and $T_1 \times T_3 = \frac{(1+A)^2 - d^2}{16}$. Since $T_2^2 \neq T_1 T_3$,they are not in $G.P.$
Checking for $A.P.$: $T_2 - T_1 = \frac{1+A}{4} - \frac{1+A-d}{2} = \frac{1+A - 2 - 2A + 2d}{4} = \frac{-1-A+2d}{4}$.
$T_3 - T_2 = \frac{1+A+d}{8} - \frac{1+A}{4} = \frac{1+A+d - 2 - 2A}{8} = \frac{-1-A+d}{8}$.
Since $T_2 - T_1 \neq T_3 - T_2$,they are not in $A.P.$
However,the sequence is of the form $\frac{a_n}{2^n}$ where $a_n$ is in $A.P.$,which defines an $A.G.P.$

(B) Let $P = a^{\frac{1}{a}} \cdot (2a)^{\frac{1}{2a}} \cdot (4a)^{\frac{1}{4a}} \cdot (8a)^{\frac{1}{8a}} \cdots \infty$.
Given $\sqrt{P} = \frac{8}{27}$,so $P = \left(\frac{8}{27}\right)^2 = \frac{64}{729}$.
We can write $P = a^{(\frac{1}{a} + \frac{1}{2a} + \frac{1}{4a} + \cdots)} \cdot 2^{(\frac{1}{2a} + \frac{2}{4a} + \frac{3}{8a} + \cdots)}$.
The exponent of $a$ is $\frac{1}{a}(1 + \frac{1}{2} + \frac{1}{4} + \cdots) = \frac{1}{a} \cdot \frac{1}{1 - 1/2} = \frac{2}{a}$.
The exponent of $2$ is an Arithmetico-Geometric Progression $(AGP)$: $S = \frac{1}{2a} + \frac{2}{4a} + \frac{3}{8a} + \cdots$.
$S = \frac{1}{a}(\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \cdots)$. Using the formula for $AGP$,$S = \frac{1}{a} \cdot \frac{1/2}{(1-1/2)^2} = \frac{1}{a} \cdot \frac{1/2}{1/4} = \frac{2}{a}$.
Thus,$P = a^{2/a} \cdot 2^{2/a} = (2a)^{2/a}$.
We have $(2a)^{2/a} = \frac{64}{729} = (\frac{8}{27})^2 = ((\frac{2}{3})^3)^2 = (\frac{2}{3})^6$.
Comparing $(2a)^{2/a} = (\frac{2}{3})^6$,we set $\frac{2}{a} = 6 \Rightarrow a = \frac{1}{3}$.
Checking the base: $2a = 2(\frac{1}{3}) = \frac{2}{3}$. This satisfies the equation.

(A) Let $S = \sum\limits_{k = 1}^{20} {\frac{k}{{{2^k}}}} = \frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} + \dots + \frac{20}{{{2^{20}}}}$
Multiply by $\frac{1}{2}$:
$\frac{1}{2}S = \frac{1}{{{2^2}}} + \frac{2}{{{2^3}}} + \dots + \frac{19}{{{2^{20}}}} + \frac{20}{{{2^{21}}}}$
Subtract the two equations:
$S - \frac{1}{2}S = \frac{1}{2} + \left( \frac{2-1}{{{2^2}}} \right) + \left( \frac{3-2}{{{2^3}}} \right) + \dots + \left( \frac{20-19}{{{2^{20}}}} \right) - \frac{20}{{{2^{21}}}}$
$\frac{1}{2}S = \left( \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \dots + \frac{1}{{{2^{20}}}} \right) - \frac{20}{{{2^{21}}}}$
The sum inside the parenthesis is a geometric progression with $a = \frac{1}{2}$,$r = \frac{1}{2}$,and $n = 20$:
Sum $= \frac{\frac{1}{2}(1 - (\frac{1}{2})^{20})}{1 - \frac{1}{2}} = 1 - \frac{1}{{{2^{20}}}}$
So,$\frac{1}{2}S = 1 - \frac{1}{{{2^{20}}}} - \frac{20}{{{2^{21}}}} = 1 - \frac{2}{{{2^{21}}}} - \frac{20}{{{2^{21}}}} = 1 - \frac{22}{{{2^{21}}}} = 1 - \frac{11}{{{2^{20}}}}$
$S = 2 - \frac{11}{{{2^{19}}}}$

(B) Given $S = \frac{7}{5} + \frac{9}{5^{2}} + \frac{13}{5^{3}} + \frac{19}{5^{4}} + \ldots$ $(1)$
Multiply by $\frac{1}{5}$:
$\frac{1}{5} S = \frac{7}{5^{2}} + \frac{9}{5^{3}} + \frac{13}{5^{4}} + \ldots$ $(2)$
Subtracting $(2)$ from $(1)$:
$S - \frac{1}{5} S = \frac{7}{5} + \left(\frac{9-7}{5^{2}}\right) + \left(\frac{13-9}{5^{3}}\right) + \left(\frac{19-13}{5^{4}}\right) + \ldots$
$\frac{4}{5} S = \frac{7}{5} + \frac{2}{5^{2}} + \frac{4}{5^{3}} + \frac{6}{5^{4}} + \ldots$
Let $T = \frac{2}{5^{2}} + \frac{4}{5^{3}} + \frac{6}{5^{4}} + \ldots$
Then $\frac{1}{5} T = \frac{2}{5^{3}} + \frac{4}{5^{4}} + \ldots$
Subtracting: $\frac{4}{5} T = \frac{2}{5^{2}} + \frac{2}{5^{3}} + \frac{2}{5^{4}} + \ldots = \frac{2/25}{1 - 1/5} = \frac{2/25}{4/5} = \frac{1}{10}$
So $T = \frac{5}{4} \times \frac{1}{10} = \frac{1}{8}$
$\frac{4}{5} S = \frac{7}{5} + \frac{1}{8} = \frac{56 + 5}{40} = \frac{61}{40}$
$S = \frac{61}{40} \times \frac{5}{4} = \frac{61}{32}$
$160 \,S = 160 \times \frac{61}{32} = 5 \times 61 = 305$

(B) Let $S = 1 \cdot 3^{0} + 2 \cdot 3^{1} + 3 \cdot 3^{2} + \dots + 10 \cdot 3^{9}$.
Multiplying by $3$,we get $3S = 1 \cdot 3^{1} + 2 \cdot 3^{2} + \dots + 9 \cdot 3^{9} + 10 \cdot 3^{10}$.
Subtracting the two equations: $S - 3S = 1 \cdot 3^{0} + (2-1) \cdot 3^{1} + (3-2) \cdot 3^{2} + \dots + (10-9) \cdot 3^{9} - 10 \cdot 3^{10}$.
$-2S = (1 + 3^{1} + 3^{2} + \dots + 3^{9}) - 10 \cdot 3^{10}$.
The sum in the parenthesis is a geometric progression with $a=1$,$r=3$,and $n=10$ terms.
$-2S = \frac{1(3^{10} - 1)}{3 - 1} - 10 \cdot 3^{10}$.
$-2S = \frac{3^{10} - 1}{2} - 10 \cdot 3^{10}$.
$-2S = \frac{3^{10} - 1 - 20 \cdot 3^{10}}{2} = \frac{-19 \cdot 3^{10} - 1}{2}$.
$S = \frac{19 \cdot 3^{10} + 1}{4}$.

(C) Given $S = 2 + \frac{6}{7} + \frac{12}{7^{2}} + \frac{20}{7^{3}} + \frac{30}{7^{4}} + \ldots$ $(1)$
Divide by $7$: $\frac{S}{7} = \frac{2}{7} + \frac{6}{7^{2}} + \frac{12}{7^{3}} + \frac{20}{7^{4}} + \ldots$ $(2)$
Subtracting $(2)$ from $(1)$:
$S - \frac{S}{7} = 2 + \left(\frac{6-2}{7}\right) + \left(\frac{12-6}{7^{2}}\right) + \left(\frac{20-12}{7^{3}}\right) + \ldots$
$\frac{6S}{7} = 2 + \frac{4}{7} + \frac{6}{7^{2}} + \frac{8}{7^{3}} + \ldots$ $(3)$
Divide $(3)$ by $7$: $\frac{6S}{49} = \frac{2}{7} + \frac{4}{7^{2}} + \frac{6}{7^{3}} + \ldots$ $(4)$
Subtracting $(4)$ from $(3)$:
$\frac{6S}{7} - \frac{6S}{49} = 2 + \left(\frac{4-2}{7}\right) + \left(\frac{6-4}{7^{2}}\right) + \left(\frac{8-6}{7^{3}}\right) + \ldots$
$\frac{42S - 6S}{49} = 2 + \frac{2}{7} + \frac{2}{7^{2}} + \frac{2}{7^{3}} + \ldots$
$\frac{36S}{49} = 2 + \left(\frac{2/7}{1 - 1/7}\right) = 2 + \left(\frac{2/7}{6/7}\right) = 2 + \frac{1}{3} = \frac{7}{3}$
$S = \frac{7}{3} \times \frac{49}{36} = \frac{343}{108}$
$4S = 4 \times \frac{343}{108} = \frac{343}{27} = \left(\frac{7}{3}\right)^{3}$

(C) We use the method of finite differences to determine the degree of the polynomial $p(x)$.
Let the values of $p(x)$ for $x = -2, -1, 0, 1, 2, 3$ be $y_i$:
$x = -2, y = -15$
$x = -1, y = 1$
$x = 0, y = 7$
$x = 1, y = 9$
$x = 2, y = 13$
$x = 3, y = 25$
First differences $(\Delta y)$: $1 - (-15) = 16, 7 - 1 = 6, 9 - 7 = 2, 13 - 9 = 4, 25 - 13 = 12$
Second differences $(\Delta^2 y)$: $6 - 16 = -10, 2 - 6 = -4, 4 - 2 = 2, 12 - 4 = 8$
Third differences $(\Delta^3 y)$: $-4 - (-10) = 6, 2 - (-4) = 6, 8 - 2 = 6$
Since the third differences are constant $(6)$,the polynomial $p(x)$ is of degree $n = 3$.
Thus,the smallest possible value of $n$ is $3$.

(C) Let $S = (20)^{19} + 2(21)(20)^{18} + 3(21)^2(20)^{17} + \ldots + 20(21)^{19}$.
Dividing by $(20)^{19}$,we get $k = 1 + 2(\frac{21}{20}) + 3(\frac{21}{20})^2 + \ldots + 20(\frac{21}{20})^{19}$.
Let $x = \frac{21}{20}$. Then $k = 1 + 2x + 3x^2 + \ldots + 20x^{19}$.
Multiplying by $x$,we get $kx = x + 2x^2 + 3x^3 + \ldots + 20x^{20}$.
Subtracting the two equations: $k(1 - x) = 1 + x + x^2 + \ldots + x^{19} - 20x^{20}$.
Using the sum of a geometric progression: $k(1 - x) = \frac{1 - x^{20}}{1 - x} - 20x^{20}$.
Since $1 - x = 1 - \frac{21}{20} = -\frac{1}{20}$,we have $k(-\frac{1}{20}) = \frac{1 - x^{20}}{-1/20} - 20x^{20} = -20(1 - x^{20}) - 20x^{20}$.
$k(-\frac{1}{20}) = -20 + 20x^{20} - 20x^{20} = -20$.
Therefore,$k = (-20) \times (-20) = 400$.

(C) Let the terms of the arithmetico-geometric progression be of the form $(a+nd)r^n$. Given the $3^{rd}$ term is $2$ and the common ratio $r=2$,we can write the terms as:
$a_1 = \frac{a-2d}{4}, a_2 = \frac{a-d}{2}, a_3 = a, a_4 = 2(a+d), a_5 = 4(a+2d)$.
Since the $3^{rd}$ term is $2$,we have $a=2$.
The terms are: $\frac{2-2d}{4}, \frac{2-d}{2}, 2, 2(2+d), 4(2+2d)$.
Sum of the $5$ terms is: $\frac{2-2d}{4} + \frac{2-d}{2} + 2 + 4+2d + 8+8d = \frac{49}{2}$.
$\frac{1-d}{2} + 1 - \frac{d}{2} + 14 + 10d = \frac{49}{2}$.
$1 - d + 14 + 10d = \frac{49}{2}$.
$15 + 9d = 24.5$.
$9d = 9.5$.
Wait,re-evaluating: The terms are $(a)r^0, (a+d)r^1, (a+2d)r^2, (a+3d)r^3, (a+4d)r^4$.
Given $a_3 = (a+2d)r^2 = 2$. With $r=2$,$(a+2d)4 = 2 \Rightarrow a+2d = 0.5$.
Sum $S_5 = a + (a+d)2 + (a+2d)4 + (a+3d)8 + (a+4d)16 = 49/2$.
$a + 2a + 2d + 4a + 8d + 8a + 24d + 16a + 64d = 24.5$.
$31a + 98d = 24.5$.
Solving $a+2d=0.5$ and $31a+98d=24.5$:
$a = 0.5 - 2d$.
$31(0.5 - 2d) + 98d = 24.5$ $\Rightarrow 15.5 - 62d + 98d = 24.5$ $\Rightarrow 36d = 9$ $\Rightarrow d = 0.25$.
$a = 0.5 - 2(0.25) = 0$.
$a_4 = (a+3d)r^3 = (0 + 3(0.25)) \times 2^3 = 0.75 \times 8 = 6$.
Re-checking the provided solution logic: The terms are $\frac{a-2d}{4}, \frac{a-d}{2}, a, 2(a+d), 4(a+2d)$.
Sum: $\frac{a}{4} - \frac{d}{2} + \frac{a}{2} - \frac{d}{2} + a + 2a + 2d + 4a + 8d = \frac{49}{2}$.
$(0.25+0.5+1+2+4)a + (-0.5-0.5+2+8)d = 24.5$.
$7.75a + 9d = 24.5$.
Given $a=2$,$7.75(2) + 9d = 24.5$ $\Rightarrow 15.5 + 9d = 24.5$ $\Rightarrow 9d = 9$ $\Rightarrow d=1$.
$a_4 = 2(a+d) = 2(2+1) = 6$.
There is a discrepancy in the provided solution's final step. Based on the sequence $a_1, a_2, 2, a_3, a_4$,$a_4$ is the $5^{th}$ term.
$a_5 = 4(a+2d) = 4(2+2) = 16$.

(B) Given $S = 109 + \frac{108}{5} + \frac{107}{5^2} + \ldots + \frac{1}{5^{108}}$.
Multiply by $\frac{1}{5}$: $\frac{S}{5} = \frac{109}{5} + \frac{108}{5^2} + \ldots + \frac{2}{5^{108}} + \frac{1}{5^{109}}$.
Subtracting the two equations:
$S - \frac{S}{5} = 109 + (\frac{108-109}{5}) + (\frac{107-108}{5^2}) + \ldots + (\frac{1-2}{5^{108}}) - \frac{1}{5^{109}}$.
$\frac{4S}{5} = 109 - (\frac{1}{5} + \frac{1}{5^2} + \ldots + \frac{1}{5^{108}}) - \frac{1}{5^{109}}$.
$\frac{4S}{5} = 109 - [\frac{1}{5} \frac{(1 - (1/5)^{108})}{(1 - 1/5)}] - \frac{1}{5^{109}}$.
$\frac{4S}{5} = 109 - [\frac{1}{5} \cdot \frac{5}{4} (1 - \frac{1}{5^{108}})] - \frac{1}{5^{109}}$.
$\frac{4S}{5} = 109 - \frac{1}{4} (1 - \frac{1}{5^{108}}) - \frac{1}{5^{109}}$.
Multiply by $\frac{5}{4}$:
$S = \frac{5}{4} [109 - \frac{1}{4} + \frac{1}{4 \cdot 5^{108}} - \frac{1}{5^{109}}]$.
$16S = 20 \cdot 109 - 5 + \frac{5}{5^{108}} - \frac{4}{5^{108}} = 2180 - 5 + \frac{1}{5^{108}}$.
Since $(25)^{-54} = (5^2)^{-54} = 5^{-108} = \frac{1}{5^{108}}$.
$16S - (25)^{-54} = 2175 + \frac{1}{5^{108}} - \frac{1}{5^{108}} = 2175$.

(A) Let $S = 1 + \frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \ldots = 10$.
Subtracting $1$ from both sides,we get $\frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \ldots = 9$.
Let $S_1 = \frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \ldots = 9$.
Then $\frac{S_1}{k} = \frac{4}{k^2} + \frac{8}{k^3} + \frac{13}{k^4} + \ldots$.
Subtracting these: $S_1(1 - \frac{1}{k}) = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots = 9(1 - \frac{1}{k})$.
Let $S_2 = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots$.
Then $\frac{S_2}{k} = \frac{4}{k^2} + \frac{4}{k^3} + \frac{5}{k^4} + \ldots$.
Subtracting these: $S_2(1 - \frac{1}{k}) = \frac{4}{k} + \frac{1}{k^3} + \frac{1}{k^4} + \ldots = \frac{4}{k} + \frac{1/k^3}{1 - 1/k} = \frac{4}{k} + \frac{1}{k^2(k-1)}$.
Substituting $S_2 = 9(1 - \frac{1}{k})$,we get $9(1 - \frac{1}{k})^2 = \frac{4}{k} + \frac{1}{k^2(k-1)}$.
$9(\frac{k-1}{k})^2 = \frac{4k(k-1) + 1}{k^2(k-1)}$.
$9(k-1)^3 = 4k^2 - 4k + 1 = (2k-1)^2$.
Testing $k=2$: $9(2-1)^3 = 9(1) = 9$,and $(2(2)-1)^2 = 3^2 = 9$.
Thus,$k=2$ is the solution.

(A) The given series is an Arithmetico-Geometric Progression $(AGP)$ of the form $\sum_{n=0}^{\infty} (a + np)r^n$,where $a = 3$,$d = p$,and $r = \frac{1}{4}$.
The sum of an infinite $AGP$ is given by $S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$.
Substituting the values into the formula:
$8 = \frac{3}{1 - \frac{1}{4}} + \frac{p \cdot \frac{1}{4}}{(1 - \frac{1}{4})^2}$
$8 = \frac{3}{\frac{3}{4}} + \frac{\frac{p}{4}}{(\frac{3}{4})^2}$
$8 = 4 + \frac{\frac{p}{4}}{\frac{9}{16}}$
$8 - 4 = \frac{p}{4} \cdot \frac{16}{9}$
$4 = \frac{4p}{9}$
$p = 9$.

(C) Let $y = 1+x$. Then $S(x) = y + 2y^2 + 3y^3 + \ldots + 60y^{60}$.
This is an arithmetico-geometric series.
$yS = y^2 + 2y^3 + \ldots + 59y^{60} + 60y^{61}$.
Subtracting the two equations: $(1-y)S = y + y^2 + y^3 + \ldots + y^{60} - 60y^{61}$.
Since $y = 1+x$,$1-y = -x$.
$-xS = \frac{y(y^{60}-1)}{y-1} - 60y^{61} = \frac{(1+x)((1+x)^{60}-1)}{x} - 60(1+x)^{61}$.
For $x=60$,$y=61$:
$-60S(60) = \frac{61(61^{60}-1)}{60} - 60(61)^{61}$.
Multiply by $-60$:
$3600 S(60) = (60)^2 S(60) = 60(61)^{61} - 61(61^{60}-1) = 60(61)^{61} - 61^{61} + 61 = 59(61)^{61} + 61$.
Comparing with $a(b)^b + b$,we get $a=59$ and $b=61$.
Thus,$a+b = 59+61 = 120$.

(C) Let $S = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \dots \infty$.
Given $S = 7$.
Multiply by $\frac{1}{7}$: $\frac{1}{7}S = \frac{1}{7}(5) + \frac{1}{7^2}(5 + \alpha) + \frac{1}{7^3}(5 + 2\alpha) + \dots \infty$.
Subtracting the two equations:
$S - \frac{1}{7}S = 5 + [\frac{1}{7}(5 + \alpha - 5) + \frac{1}{7^2}(5 + 2\alpha - (5 + \alpha)) + \dots \infty]$.
$\frac{6}{7}S = 5 + [\frac{\alpha}{7} + \frac{\alpha}{7^2} + \frac{\alpha}{7^3} + \dots \infty]$.
The term in the bracket is an infinite geometric series with first term $a = \frac{\alpha}{7}$ and common ratio $r = \frac{1}{7}$.
Sum $= \frac{a}{1-r} = \frac{\alpha/7}{1 - 1/7} = \frac{\alpha/7}{6/7} = \frac{\alpha}{6}$.
So,$\frac{6}{7}S = 5 + \frac{\alpha}{6}$.
Since $S = 7$,we have $\frac{6}{7}(7) = 5 + \frac{\alpha}{6}$.
$6 = 5 + \frac{\alpha}{6}$ $\Rightarrow 1 = \frac{\alpha}{6}$ $\Rightarrow \alpha = 6$.

(C) The forward difference operator $\Delta$ is defined as $\Delta u_{n} = u_{n+1} - u_{n}$.
For the third-order forward difference,we use the formula $\Delta^{3} u_{0} = (E-1)^{3} u_{0}$,where $E$ is the shift operator such that $E u_{n} = u_{n+1}$.
Expanding the operator,we get $\Delta^{3} u_{0} = (E^{3} - 3E^{2} + 3E - 1) u_{0}$.
This simplifies to $\Delta^{3} u_{0} = u_{3} - 3u_{2} + 3u_{1} - u_{0}$.
Substituting the given values $u_{0}=8, u_{1}=3, u_{2}=12, u_{3}=51$:
$\Delta^{3} u_{0} = 51 - 3(12) + 3(3) - 8$.
$\Delta^{3} u_{0} = 51 - 36 + 9 - 8$.
$\Delta^{3} u_{0} = 15 + 9 - 8 = 24 - 8 = 16$.

(D) To find $\Delta^{5} u_{x}$,we construct the forward difference table:

$x, u_{x}$	$\Delta u_{x}, \Delta^{2} u_{x}, \Delta^{3} u_{x}, \Delta^{4} u_{x}, \Delta^{5} u_{x}$
$0, 3$	$9, 60, -10, -259, 755$
$1, 12$	$69, 50, -269, 496$
$2, 81$	$119, -219, 227$
$3, 200$	$-100, 8$
$4, 100$	$-92$
$5, 8$	-

The forward difference $\Delta^{5} u_{0}$ is calculated as follows:
$\Delta u_{0} = u_{1} - u_{0} = 12 - 3 = 9$
$\Delta u_{1} = u_{2} - u_{1} = 81 - 12 = 69$
$\Delta u_{2} = u_{3} - u_{2} = 200 - 81 = 119$
$\Delta u_{3} = u_{4} - u_{3} = 100 - 200 = -100$
$\Delta u_{4} = u_{5} - u_{4} = 8 - 100 = -92$
Continuing this process for higher-order differences,we obtain the final value $\Delta^{5} u_{0} = 755$.

(B) The given series is $1 + \frac{3}{7} + \frac{5}{7^2} + \frac{7}{7^3} + \dots$
The numerators are $1, 3, 5, 7, \dots$,which form an Arithmetic Progression $(AP)$ with first term $a = 1$ and common difference $d = 2$.
The $n$th term of this $AP$ is $T_n(AP) = a + (n-1)d = 1 + (n-1)2 = 2n - 1$.
The denominators are $7^0, 7^1, 7^2, 7^3, \dots$,which form a Geometric Progression $(GP)$ with first term $a = 1$ and common ratio $r = 7$.
The $n$th term of this $GP$ is $T_n(GP) = ar^{n-1} = 1 \times 7^{n-1} = 7^{n-1}$.
Therefore,the $n$th term of the given series is $T_n = \frac{2n-1}{7^{n-1}}$.