====== Question 2 & 3 Exercise 5.2 ======
Solutions of Question 2 & 3 of Exercise 5.2 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 2=====
$1+3^2 x+5^2 x^2+7^2 x^3+\ldots, x<1$.
====Solution====
Let
\begin{align}
& S_{\infty}=1+3^2 x+5^2 x^2+7^2 x^3+\ldots ..(1)\\
& x S_{\infty}=x+3^2 x^2+5^2 x^3+7^2 x^4+\ldots..(2)\end{align}
Subtracting the (2) from (2), we get
\begin{align}& (1-x) S_{\infty}=1^2+(3^2-1^2) x+(5^2-3^2) x^2+(7^2-5^2) x^3+\ldots \\
& (1-x) S_{\infty}=1+8 x+16 x^2+24 x^3+\ldots \quad \text { (3) } \\
& \Rightarrow x(1-x) S_{\infty}=x+8 x^2+16 x^3+24 x^4+\ldots(4)\end{align}
Again subtracting (4) from (3)
\begin{align}
& [(1-x)-x(1-x)] S_{\infty}=1+(8-1) x+(16-8) x^2+(24-16) x^3+\ldots \\
& \Rightarrow(1-2 x+x^2) S_{\infty}=1+7 x+ 8(x^2+x^3+x^4+\ldots) \\
& \Rightarrow(x-1)^2 S_{\infty}=1+7 x+8\dfrac{x^2}{1-x} \\
& \Rightarrow S_{\infty}=\dfrac{1+7 x}{(x-1)^2}-\dfrac{8x^2}{(x-1)^3} \end{align}

=====Question 3=====
Find the $n^{\text {th }}$ term of the following arithmetic-geometric series:
$\dfrac{0}{1}+\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\dfrac{4}{16}+\dfrac{5}{32}+\ldots$
====Solution====
The given series is the product of the corresponding terms of the series: $0,1.2,3, \ldots$ and $1, \dfrac{1}{2}, \dfrac{1}{4}.$

The $nth$ term of the the arithmetic series is: $$a_n=n$$ 
$nth$ term of the geometric series is $$b_n=(\dfrac{1}{2})^{n-1}$$
Thus the $n^{\text {th }}$ term of the given arithmetic-geometric series is:
\begin{align}
& T_n=a_n \times b_n \\
& T_n=(n)\cdot(\dfrac{1}{2})^{n-1}\end{align}


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