Question

If $x = \frac{3}{10} + \frac{3.7}{10.15} + \frac{3.7.9}{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

The Correct Option is D

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