Question

Find the sum of the series upto infinity.
(x+y) + (x^2+xy+y^2) + (x^3+x^2y+xy^2+y^3) + . . . . .


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Answer 1

Pre-requisite Knowledge :-

• $\rm x^2 - y^2 = ( x-y ) ( x^2+xy+y^2 )$

• $\rm x^3 - y^3 = ( x-y ) ( x^3+x^2y+xy^2+y^3)$

• $\rm x^4-y^4=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$

This above series will continue upto infinity.

Now let's solve the given problem!

Given that,

$\rm\longrightarrow S_n = ( x + y ) + ( x^2 + xy + y^2 ) + ( x^3 + x^2 y + xy^2 + y^3)+\cdots\cdots\infty$

Multiplying numerator and denominator with $\rm(x-y)$

$\rm\longrightarrow S_n = \dfrac{1}{(x-y)}\bigg[(x-y)(x + y )+ (x-y)( x^2 + xy + y^2 ) + (x-y)( x^3 + x^2 y + xy^2 + y^3)+\cdots\cdots\infty \bigg]$

$\rm\longrightarrow S_n = \dfrac{1}{(x-y)}\bigg[(x^2 - y^2)+ (x^3 - y^3)+(x^3-y^4)+\cdots\cdots\infty \bigg]$

$\rm\longrightarrow S_n = \dfrac{1}{(x-y)}\bigg[(x^2 + x^3+x^4+...\infty)-(y^2+y^3+y^4 +\cdots\infty) \bigg]$

Now, we can see that $\rm(x^2 + x^3 + x^4 + ...\infty )$ and $\rm(y^2 + y^3 + y^4 + ...\infty )$ are forming infinite G.P., We can use the below formula to find the sum of infinite G.P.

Sum of infinite GP is given by :-

• $\tt\red{ S_\infty = \dfrac{a}{1-r}, where \,a = First \,term , \,r=common \,ratio}$

Using this formula,

$\rm\longrightarrow S_n = \dfrac{1}{(x-y)}\bigg[(x^2 + x^3+x^4+...\infty)-(y^2+y^3+y^4 +\cdots\infty) \bigg]$

$\rm\longrightarrow S_n = \dfrac{1}{(x-y)}\bigg[\dfrac{x^2}{1-x}-\dfrac{y^2}{1-y}\bigg]$

$\rm\longrightarrow S_n = \dfrac{1}{(x-y)}\bigg[\dfrac{x^2(1-y)-y^2(1-x)}{(1-x)(1-y)}\bigg]$

$\rm\longrightarrow S_n = \dfrac{1}{(x-y)}\bigg[\dfrac{x^2-x^2y-y^2+y^2x}{(1-x)(1-y)}\bigg]$

$\rm\longrightarrow S_n = \dfrac{1}{(x-y)}\bigg[\dfrac{x^2-y^2-x^2y+y^2x}{(1-x)(1-y)}\bigg]$

$\rm\longrightarrow S_n = \dfrac{1}{(x-y)}\bigg[\dfrac{(x+y)(x-y)+(-xy)(x-y)}{(1-x)(1-y)}\bigg]$

$\rm\longrightarrow S_n = \dfrac{1}{(x-y)}\left[\dfrac{(x-y)\Big[(x+y)+(-xy)\Big]}{(1-x)(1-y)}\right]$

$\rm\longrightarrow S_n = \dfrac{(x-y)}{(x-y)}\left[\dfrac{\Big[(x+y)+(-xy)\Big]}{(1-x)(1-y)}\right]$

$\rm\longrightarrow S_n = 1\left[\dfrac{\Big[(x+y)+(-xy)\Big]}{(1-x)(1-y)}\right]$

$\rm\longrightarrow S_n = \left[\dfrac{(x+y-xy)}{(1-x)(1-y)}\right]$

$\rm\longrightarrow S_n = \left[\dfrac{x+y-xy}{(1-x)(1-y)}\right]$

This is the required answer.