$\frac{1}{4}-\frac{5}{4 \cdot 8}+\frac{5 \cdot 7}{4 \cdot 8 \cdot 12}-\ldots=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext {Let } S=\frac{1}{4}-\frac{5}{4 \cdot 8}+\frac{5 \cdot 7}{4 \cdot 8 \cdot 12}-\ldots \ & =\frac{1}{3}\left(\frac{3}{4}-\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}-\ldotsight)\end{aligned}$ Now, we know that $ (1+x)^{-n}=1-n x+\frac{n(n+1)}{2 !} x^2 \ldots \ldots $ $ \begin{aligned} & 	ext { Put } n=\frac{3}{2} 	ext { and } n=\frac{1}{2} \ & \begin{aligned} \Rightarrow \quad \frac{3}{2} \cdot \frac{1}{2} & -\frac{\frac{3}{2} \cdot \frac{5}{2}}{2 !}\left(\frac{1}{2}ight)^2+\ldots \ = & 1-\left(1+\frac{1}{2}ight)^{-3 / 2}=1-\left(\frac{3}{2}ight)^{-3 / 2} \ & =1-\left(\frac{2}{3}ight)^{3 / 2}=1-\frac{2 \sqrt{2}}{3 \sqrt{3}}=\frac{3 \sqrt{3}-2 \sqrt{2}}{3 \sqrt{3}} \end{aligned} \end{aligned} $ put in Eq. (i), we get $ S=\frac{1}{3}\left(\frac{3 \sqrt{3}-2 \sqrt{2}}{3 \sqrt{3}}ight) \Rightarrow S=\frac{3 \sqrt{3}-2 \sqrt{2}}{9 \sqrt{3}} $

$\frac{1}{4}-\frac{5}{4 \cdot 8}+\frac{5 \cdot 7}{4 \cdot 8 \cdot 12}-\ldots=$

Solution