Question

If $x=\frac{3}{10}+\frac{3.7}{10.15}+\frac{\mathrm{3.7.9}}{\mathrm{10.15.20}}+$ ...., then $5x+8$ =

• A. $\frac{5\sqrt{5}}{3\sqrt{3}}$
• B. $\frac{5\sqrt{5}}{\sqrt{3}}$
• C. $\frac{3\sqrt{3}}{\sqrt{5}}$
• D. $\frac{25\sqrt{5}}{3\sqrt{3}}$

Answer 1

Answer: (D)

Given,
$x=\frac{3}{10}+\frac{3\cdot 7}{10\cdot 15}+\frac{3\cdot 7\cdot 9}{10\cdot 15\cdot 20}+\dots$
$x=\frac{3\cdot 5}{5\cdot 10}+\frac{3\cdot 5\cdot 7}{5\cdot 10\cdot 15}+\frac{3\cdot 5\cdot 7\cdot 9}{5\cdot 10\cdot 15\cdot 20}+\dots$
$\therefore T_{r+1}=\frac{3\cdot 5\cdot 7\dots (2r+1)}{5^r\cdot 1\cdot 2\cdot 3\dots r}$
$={\left(\frac{2}{5}\right)}^r\frac{3/2\cdot 5/2\cdot \frac{7}{2}\dots \left(r+\frac{1}{2}\right)}{r!}$
$=\frac{\left(-\frac{3}{2}\right)\left(-\frac{3}{2}-1\right)\left(-\frac{3}{2}-2\right)\dots \left(-\frac{3}{2}-r+1\right){\left(-\frac{2}{5}\right)}^r}{r!}$
Comparing with general term of $(1+x{)}^n,n\in R$
$\therefore \frac{n(n-1)(n-2)\dots (n-r+1)x^r}{r!}$
$=\frac{\left(-\frac{3}{2}\right)\left(-\frac{3}{2}-1\right)\dots \left(-\frac{3}{2}-r+1\right){\left(-\frac{2}{5}\right)}^r}{r!}$
$\Rightarrow n=-\frac{3}{2^{\prime }}x=-\frac{2}{5}$
$\therefore x+\frac{8}{5}={\left(1-\frac{2}{5}\right)}^{-3/2}$
$={\left(\frac{3}{5}\right)}^{-3/2}={\left(\frac{5}{3}\right)}^{3/2}$
$=\frac{5\sqrt{5}}{3\sqrt{3}}$
$\Rightarrow \frac{5x+8}{5}=\frac{5\sqrt{5}}{3\sqrt{3}}$
$\Rightarrow 5x+8=\frac{25\sqrt{5}}{3\sqrt{3}}$