Question

Find the sum of the infinite series $1+\frac{2}{3}.\frac{1}{2}+\frac{2.5}{3.6}(\frac{1}{2}{)}^2+\frac{\mathrm{2.5.8}}{\mathrm{3.6.9}}(\frac{1}{2}{)}^3+....\infty$

Answer 1

The given infinite series is,

$S=1+\frac{2}{3}.\frac{1}{2}+\frac{2}{3}.\frac{5}{6}.{\left(\frac{1}{2}\right)}^2+\frac{2}{3}.\frac{5}{6}.\frac{8}{9}.{\left(\frac{1}{2}\right)}^3------------$ (1)

Now, expansion of binomial series ${(1+x)}^n$ is given by,

${(1+x)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^2+----------$ (2)

Comparing equation (1) and (2), first term of both equations is same.

Thus, comparing second term of both equations, we get,

$nx=\frac{2}{3}\times \frac{1}{2}$

$\therefore nx=\frac{1}{3}$

Squaring both sides, we get,

$\therefore n^2{x}^2=\frac{1}{9}$

$\therefore x^2=\frac{1}{9n^2}$ (3)

Comparing third term of both series, we get,

$\frac{n(n-1)}{2!}{x}^2=\frac{2}{3}.\frac{5}{6}.{\left(\frac{1}{2}\right)}^2$

$\therefore \frac{n(n-1)}{2}{x}^2=\frac{5}{9}.\frac{1}{4}$

$\therefore \frac{n(n-1)}{2}{x}^2=\frac{5}{36}$

From equation (3),

$\therefore \frac{n(n-1)}{2}.\frac{1}{9n^2}=\frac{5}{36}$

$\therefore n(n-1)\times \frac{1}{n^2}=\frac{5\times 18}{36}$

$\therefore \frac{(n-1)}{n}=\frac{5}{2}$

$\therefore (n-1)=\frac{5}{2}n$

$\therefore \frac{5}{2}n-n=-1$

$\therefore \frac{3}{2}n=-1$

$\therefore n=\frac{-2}{3}$

From equation (1),

$\frac{-2}{3}x=\frac{1}{3}$

$\therefore x=\frac{-1}{2}$

Thus, sum of series is given by,

$S={(1+x)}^n$

$S={(1-\frac{1}{2})}^{\frac{-2}{3}}$

$S={\left(\frac{1}{2}\right)}^{\frac{-2}{3}}$

$\therefore S=\frac{1}{{\left(2\right)}^{\frac{-2}{3}}}$

$\therefore S={\left(2\right)}^{\frac{2}{3}}$

$\therefore S={\left(4\right)}^{\frac{1}{3}}$