Find the sum of the GP :

Question:

Find the sum of the GP :

$x(x+y)+x^{2}\left(x^{2}+y^{2}\right)+x^{3}\left(x^{3}+y^{3}\right)+\ldots .$ To $n$ terms

 

Solution:

The given expression can be written as

$=\left(x^{2}+x y\right)+\left(x^{4}+x^{2} y^{2}\right)+\left(x^{6}+x^{3} y^{3}\right)+\ldots$ To $n$ terms

$=\left(x^{2}+x^{4}+x^{6}+\ldots\right.$ to $n$ terms $)+\left(x y+x^{2} y^{2}+x^{3} y^{3}+\ldots\right.$ to $n$ terms $)$

Sum of a G.P. series is represented by the formula $S_{n}=a \frac{r^{n}-1}{r-1}$ when r≠1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

$a=x^{2}$ first part and $x y$ for the second part

$r=($ ratio between the $n$ term and $n-1$ term $) x^{2}$ for the first part and $x y$ for the second part

n terms

$\therefore \mathrm{S}_{\mathrm{n}}=\mathrm{x}^{2} \times \frac{\mathrm{x}^{2 \mathrm{n}}-1}{\mathrm{x}^{2}-1}+\mathrm{xy} \times \frac{\mathrm{x}^{\mathrm{n}} \mathrm{y}^{\mathrm{n}}-1}{\mathrm{xy}-1}$

$\Rightarrow S_{n}=\frac{x^{2}\left(x^{n}-1\right)\left(x^{n}+1\right)}{(x+1)(x-1)}+\frac{x^{n+1} y^{n+1}-1}{x y-1}$

 

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