If $x=\frac{3}{10}+\frac{3 \cdot 7}{10 \cdot 15}+\frac{3 \cdot 7 \cdot 9}{10 \cdot 15 \cdot 20}+\ldots$,…
If $x=\frac{3}{10}+\frac{3 \cdot 7}{10 \cdot 15}+\frac{3 \cdot 7 \cdot 9}{10 \cdot 15 \cdot 20}+\ldots$, then $5 x+8=$
- $\frac{5 \sqrt{5}}{3 \sqrt{3}}$
- $\frac{5 \sqrt{5}}{\sqrt{3}}$
- $\frac{3 \sqrt{3}}{\sqrt{5}}$
- $\frac{25 \sqrt{5}}{3 \sqrt{3}}$
Solution
Given,
$
\begin{aligned}
& 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 !}
\end{aligned}
$
Comparing with general term of $(1+x)^n, n \in R$
$
\begin{aligned}
& \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}}
\end{aligned}
$
Asked in: AP EAMCET 2019 (21 Apr Shift 1)
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