Question

please tell the complete solution of this question....it is from the topic AGP ​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Sum to infinity of the series

$\rm :\longmapsto\: {1}^{2} + {2}^{2}x + {3}^{2} {x}^{2} + {4}^{2} {x}^{3} + - - - \infty , \: |x| < 1$

$\rm :\longmapsto\:(A) \: \: \dfrac{1 + x}{ {(1 - x)}^{3} }$

$\rm :\longmapsto\:(B) \: \: \dfrac{1 - x}{ {(1 + x)}^{3} }$

$\rm :\longmapsto\:(C) \: \: \dfrac{2 + x}{ {(1 - x)}^{3} }$

$\rm :\longmapsto\:(D) \: \: none \: of \: these$

$\large\underline{\sf{Solution-}}$

Given infinite series is

$\rm :\longmapsto\: {1}^{2} + {2}^{2}x + {3}^{2} {x}^{2} + {4}^{2} {x}^{3} + - - - \infty , \: |x| < 1$

Let assume that

$\rm :\longmapsto\: S = {1}^{2} + {2}^{2}x + {3}^{2} {x}^{2} + {4}^{2} {x}^{3} + - - - \infty - - (1)$

can be rewritten as

$\rm :\longmapsto\: S = 1 + 4x + 9 {x}^{2} + 16 {x}^{3} + - - - \infty - - (2)$

Now, this sum is product of corresponding terms of two series

$\rm :\longmapsto\: {1}^{2}, {2}^{2}, {3}^{2}, {4}^{2}, - - -$

and

$\rm :\longmapsto\:1,x, {x}^{2}, {x}^{3}, - - -$

So, second series is an GP series with common ratio x.

So,

On multiply equation (2) by x, on both sides we get

$\rm :\longmapsto\: xS = x+4 {x}^{2} + 9 {x}^{3}+16 {x}^{4} + - - - \infty - - (3)$

On Subtracting equation (3) from equation (2), we get

$\rm :\longmapsto\: (1 - x)S = 1 + 3x+5 {x}^{2} + 7{x}^{3}+ - - - \infty - - (4)$

On multiply equation (4) by x, we get

$\rm :\longmapsto\: x(1 - x)S = x + 3 {x}^{2} +5 {x}^{3} + 7{x}^{4}+ - - - \infty - - (5)$

On Subtracting equation (5) from equation (4), we get

$\rm :\longmapsto\: (1 - x)(1 - x)S =1 + 2x + 2 {x}^{2} +2{x}^{3} + - - - \infty$

$\rm :\longmapsto\: {(1 - x)}^{2}S = 1 + 2x(1 + x + {x}^{2} + - - - \infty )$

We know,

Sum of infinite GP series is given by

$\boxed{ \bf{ \: S_ \infty = \frac{a}{1 - r}, \: provided \: that \: |r| < 1}}$

where,

a is first term of GP series

r is common ratio.

So, using this result,

$\rm :\longmapsto\: {(1 - x)}^{2}S = 1 + 2x\bigg[\dfrac{1}{1 - x}\bigg]$

$\rm :\longmapsto\: {(1 - x)}^{2}S = 1 + \dfrac{2x}{1 - x}$

$\rm :\longmapsto\: {(1 - x)}^{2}S = \dfrac{1 - x + 2x}{1 - x}$

$\rm :\longmapsto\: {(1 - x)}^{2}S = \dfrac{1 + x}{1 - x}$

$\bf\implies \:S = \dfrac{1 + x}{ {(1 - x)}^{3} }$

• Hence, Option A is correct.