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$1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\ldots$ is equal to
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Verified Answer
The correct answer is:
$\sqrt{2}$
We have,
$\begin{aligned} & (1-x)^{-1 / 2}=1+\frac{1}{2} x+\frac{\frac{1}{2}\left(\frac{1}{2}+1\right)}{2 !} x^2 \\ & \frac{+\frac{1}{2}\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+2\right)}{3 !} x^3+\ldots \\ & (1-x)^{-1 / 2}=1+\frac{1}{2} x+\frac{1.3}{8} x^2+\frac{1 \cdot 3 \cdot 5}{4 \cdot 12} x^3+\ldots \\ & \end{aligned}$
Put $x=\frac{1}{2}$
$\begin{gathered}\left(1-\frac{1}{2}\right)^{-1 / 2}=1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\ldots \\ \sqrt{2}=1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\ldots\end{gathered}$
$\begin{aligned} & (1-x)^{-1 / 2}=1+\frac{1}{2} x+\frac{\frac{1}{2}\left(\frac{1}{2}+1\right)}{2 !} x^2 \\ & \frac{+\frac{1}{2}\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+2\right)}{3 !} x^3+\ldots \\ & (1-x)^{-1 / 2}=1+\frac{1}{2} x+\frac{1.3}{8} x^2+\frac{1 \cdot 3 \cdot 5}{4 \cdot 12} x^3+\ldots \\ & \end{aligned}$
Put $x=\frac{1}{2}$
$\begin{gathered}\left(1-\frac{1}{2}\right)^{-1 / 2}=1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\ldots \\ \sqrt{2}=1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\ldots\end{gathered}$
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