1+frac{1}{4}+frac{1 cdot 3}{4 cdot 8}+frac{1 | vedclass

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Question

(A) The general binomial expansion for $(1-x)^{-n}$ is given by $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \ldots$ For

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$\sqrt{2}$

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(A) The general binomial expansion for $(1-x)^{-n}$ is given by $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \ldots$ For $n = \frac{1}{2}$,we have $(1-x)^{-1/2} = 1 + \frac{1}{2}x + \frac{\frac{1}{2}(\frac{3}{2})}{2!}x^2 + \frac{\frac{1}{2}(\frac{3}{2})(\frac{5}{2})}{3!}x^3 + \ldots$ $(1-x)^{-1/2} = 1 + \frac{1}{2}x + \frac{1 \cdot 3}{8}x^2 + \frac{1 \cdot 3 \cdot 5}{48}x^3 + \ldots$ Comparing this with the given series $1 + \frac{1}{4} + \frac{1 \cdot 3}{32} + \ldots$,we set $\frac{1}{2}x = \frac{1}{4}$,which gives $x = \frac{1}{2}$. Substituting $x = \frac{1}{2}$ into the expansion: $(1 - \frac{1}{2})^{-1/2} = (\frac{1}{2})^{-1/2} = \sqrt{2}$. Thus,the sum of the series is $\sqrt{2}$.

$1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\ldots$ is equal to

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