1+frac{2}{4}+frac{2 cdot 5}{4 cdot 8}+frac{2 | vedclass

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(B) Let $S=1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\ldots$Comparing this with the binomial expansion $(

Vedclass · Accepted Answer

$\sqrt[3]{16}$

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(B) Let $S=1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\ldots$Comparing this with the binomial expansion $(1-x)^n = 1 + n(-x) + \frac{n(n-1)}{2!}(-x)^2 + \ldots$We have $nx = \frac{2}{4} = \frac{1}{2}$ $(i)$And $\frac{n(n-1)}{2!}x^2 = \frac{2 \cdot 5}{4 \cdot 8} = \frac{10}{32} = \frac{5}{16}$ (ii)Dividing (ii) by the square of $(i)$: $\frac{\frac{n(n-1)}{2}x^2}{n^2x^2} = \frac{5/16}{1/4} \Rightarrow \frac{n-1}{2n} = \frac{5}{4}$$4n - 4 = 10n$ $\Rightarrow -6n = 4$ $\Rightarrow n = -2/3$Substituting $n$ into $(i)$: $(-2/3)x = 1/2 \Rightarrow x = -3/4$Thus,$S = (1 - x)^n = (1 - (-3/4))^{-2/3} = (7/4)^{-2/3}$ is incorrect. Re-evaluating: The series is $(1-x)^n$ where $nx = 1/2$ and $x = 1/2$ is not correct. Using $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \ldots$Here $nx = 1/2$ and $\frac{n(n+1)}{2}x^2 = 5/16$. Solving gives $n=2/3$ and $x=3/4$.$S = (1-3/4)^{-2/3} = (1/4)^{-2/3} = 4^{2/3} = \sqrt[3]{16}$.

$1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots$ is equal to :

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