Question

The sum to infinite terms of the series $\frac{3}{10}+\frac{3.7}{10.15}+\frac{3.7 .9}{10.15 .20}+\ldots$ 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}+\ldots $
$= \frac{3}{5 \times 2 !}+\frac{3 \cdot 7}{5^{2} \cdot 3 !}+\frac{3 \cdot 7 \cdot 9}{5^{3} \cdot 4 !}+\ldots$
We know that,
$(1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2}+\frac{n(n-1)(n-2)}{3 !} x^{3}+\ldots$
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}-4 n+4\right) $
$ \Rightarrow \, 34 n^{2}+ 11 n-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} $