Question

The sum to infinite terms of the series $\frac{3}{10}+\frac{3.7}{10.15}+\frac{\mathrm{3.7.9}}{\mathrm{10.15.20}}+\dots$ to $\infty$ is

• A. $\sqrt[4]{125}-1$
• B. $\frac{5\sqrt{5}}{3\sqrt{3}}-\frac{8}{5}$
• C. $\sqrt[3]{4}-\frac{4}{3}$
• D. $\sqrt{\frac{5}{3}}-\frac{6}{5}$

Answer 1

Answer: (B)

We have series
$\frac{3}{10}+\frac{3\cdot 7}{10\cdot 15}+\frac{3\cdot 7\cdot 9}{10\cdot 15\cdot 20}+\dots$
$=\frac{3}{5\times 2!}+\frac{3\cdot 7}{5^2\cdot 3!}+\frac{3\cdot 7\cdot 9}{5^3\cdot 4!}+\dots$
We know that,
$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+\dots$
Here comparing,
$\frac{n(n-1)}{2!}{x}^2=\frac{3}{5\times 2!}$
and $\frac{n(n-1)(n-2)}{3!}{x}^3=\frac{3\cdot 7}{5^2\times 3!}$
$n(n-1)x^2=\frac{3}{5}\dots (i)$
and $n(n-1)(n-2)x^3=\frac{21}{25}\dots (iii)$
$\Rightarrow (n-2)x\times \frac{3}{5}=\frac{21}{25}$
$\Rightarrow x=\frac{7}{5(n-2)}$
Put $x=\frac{7}{5(n-2)}$ in Eq. (i), we get
$n(n-1)\frac{49}{25(n-2{)}^2}=\frac{3}{5}$
$\Rightarrow \left(n^2-n\right)49=15\left(n^2-4n+4\right)$
$\Rightarrow 34n^2+11n-60=0$
$\Rightarrow n=-\frac{3}{2}$ or $n=\frac{20}{17}$
$\therefore x=\frac{7}{5}\frac{1}{\left(-\frac{3}{2}-2\right)}=-\frac{2}{5}$
$\therefore$ Sum of series is $={\left(1-\frac{2}{5}\right)}^{\frac{-3}{2}}-1-\frac{3}{5}$
$={\left(\frac{5}{3}\right)}^{3/2}-\frac{8}{5}=\frac{5\sqrt{5}}{3\sqrt{3}}-\frac{8}{5}$