Question

If $\left|x\right|<1$ and $\left|y\right|<1,$ find the sum to infinity of the series

$(x+y)+(x^2+xy+y^2)+(x^3+x^{2}y+x{y}^2+y^3)+....$

Or

Find the sum to n terms of the series $\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....$

Answer 1

Let $(x+y)+(x^2+xy+y^2)+(x^3+x^{2}y+x{y}^2+y^3)+...$

On multiplying and dividing by (x - y), we get

$E=\frac{(x^2-y^2)}{x-y}+\frac{x^3-y^3}{x-y}+\frac{x^4-y^4}{x-y}+...$

= $\frac{1}{x-y}\left[\right(x^2+x^3+x^4+...\infty )-(y^2+y^3+y^4+...\infty \left)\right]$

= $\frac{1}{x-y}(\frac{x^2}{1-x}-\frac{y^2}{1-y})[\because s_{\infty }=\frac{a}{1-r}]$

= $\frac{1}{x-y}\left[\frac{x^2(1-y)-y^2(1-x)}{(1-x)(1-y)}\right]$

= $\frac{1}{x-y}\left[\frac{x^2-x^{2}y-y^2+x{y}^2}{(1-x)(1-y)}\right]$

= $\frac{1}{x-y}\left[\frac{(x^2-y^2)-xy(x-y)}{(1-x)(1-y)}\right]$

= $\frac{(x-y)(x+y-xy)}{(x-y)(1-x)(1-y)}=\frac{x+y-xy}{(1-x)(1-y)}$

Or

Here, nth term of $1+3+5x...$ is

$a_n=1+(n-1)2[\because a_n=a+(n-1\left)d\right]$

= $2n-1$

Similarly, nth term of $3+5+7+...$ is

$a_n^{\prime }=2n+1$

Now, nth term of the given series is

$T_n=\frac{1}{(2n-1)(2n+1)};n=1,2,3,...,n$

= $\frac{1}{2}[\frac{1}{2n-1}-\frac{1}{2n+1}]$

$\therefore$ Required sum = $\sum T_n$

= $\sum \frac{1}{2}[\frac{1}{2n-1}-\frac{1}{2n+1}]$

= $\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+...+(\frac{1}{2n-1}-\frac{1}{2n+1})]$

= $\frac{1}{2}(1-\frac{1}{2n+1})$

= $\frac{1}{2}\left(\frac{2n+1-1}{2n+1}\right)=\frac{n}{2n+1}$

Hence, sum of the given series is $\frac{n}{2n+1}$