# Find the sum to n terms of the sequence :

Question:

Find the sum to n terms of the sequence :

(i) $\left(x+\frac{1}{x}\right)^{2},\left(x^{2}+\frac{1}{x^{2}}\right)^{2},\left(x^{3}+\frac{1}{x^{3}}\right)^{2}$ ,….. to n terms

(ii) $\left.(x+y), 9 x^{2}+x y+y^{2}\right),\left(x^{3}+x 2 y+x y^{2}+y^{3}\right), \ldots .$ to $n$ terms

Solution:

This can also be written as

$=\left(x^{2}+\frac{1}{x^{2}}+2\right)+\left(x^{4}+\frac{1}{x^{4}}+2\right)+\left(x^{6}+\frac{1}{x^{6}}+2\right)+\ldots \ldots$ to $n$ term

$\left(x^{2}+x^{4}+x^{6}+\ldots\right.$ to $n$ terms $)+\left(\frac{1}{x^{2}}+\frac{1}{x^{4}}+\frac{1}{x^{6}}+\ldots\right.$ to $n$ terms $)+(2+$

$=2+2+\ldots .$ to $\mathrm{n}$ terms)

$=\left(x^{2}+x^{4}+x^{6}+\ldots\right.$ to $n$ terms $)+\left(\frac{1}{x^{2}}+\frac{1}{x^{4}}+\frac{1}{x^{6}}+\ldots .\right.$ to $n$ terms $)+2 n$

Sum of a G.P. series is represented by the formula, $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$

hen 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.

$a=x^{2}, \frac{1}{x^{2}}$

$r=$ (ratio between the $n$ term and $n-1$ term) $x^{2}, \frac{1}{x^{2}}$

n terms

$\therefore \mathrm{S}_{\mathrm{n}}=\mathrm{x}^{2} \times \frac{\mathrm{x}^{2 \mathrm{n}}-1}{\mathrm{x}^{2}-1}+\frac{1}{\mathrm{x}^{2}} \times \frac{\left(\frac{1}{\mathrm{x}^{2}}\right)^{\mathrm{n}}-1}{\frac{1}{\mathrm{x}^{2}}-1}+2 \mathrm{n}$

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{x}^{2}\left(\mathrm{x}^{\mathrm{n}}-1\right)\left(\mathrm{x}^{\mathrm{n}}+1\right)}{(\mathrm{x}-1)(\mathrm{x}+1)}+\frac{1}{\mathrm{x}^{2}} \times \frac{\frac{1}{\mathrm{x}^{2}}^{\mathrm{n}}-1}{\frac{\mathrm{x}^{2}-1}{\mathrm{x}^{2}}}+2 \mathrm{n}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{x}^{2}\left(\mathrm{x}^{\mathrm{n}}-1\right)\left(\mathrm{x}^{\mathrm{n}}+1\right)}{(\mathrm{x}-1)(\mathrm{x}+1)}+\frac{\frac{1}{\mathrm{x}^{2}}^{\mathrm{n}}-1}{\mathrm{x}^{2}-1}+2 \mathrm{n}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{x}^{2}\left(\mathrm{x}^{\mathrm{n}}-1\right)\left(\mathrm{x}^{\mathrm{n}}+1\right)}{(\mathrm{x}-1)(\mathrm{x}+1)}+\frac{\frac{1}{\mathrm{x}^{2}}^{\mathrm{n}}-1}{(\mathrm{x}-1)(\mathrm{x}+1)}+2 \mathrm{n}$

$\therefore \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{x}^{2}\left(\mathrm{x}^{\mathrm{n}}-1\right)\left(\mathrm{x}^{\mathrm{n}}+1\right)+\frac{1}{\mathrm{x}^{2}}^{\mathrm{n}}-1}{(\mathrm{x}-1)(\mathrm{x}+1)}+2 \mathrm{n}$

(ii) If we divide and multiply the terms by (x-y)

$=\frac{(x-y)(x+y)+(x-y)\left(x^{2}+x y+y^{2}\right)+(x-y)\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots \text { to } n \text { terms }}{(x-y)}$

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

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

Sum of a G.P. series is represented by the formula, $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$

hen 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.

Here,

$a=x^{2}, y^{2}$

r = (ratio between the n term and n-1 term) x, y

n terms

$\therefore S_{n}=\frac{x^{2} \times \frac{x^{n}-1}{x-1}+y^{2} \times \frac{y^{n}-1}{y-1}}{(x-y)}$

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