The sum of the series 1 + frac{1 times 3}{6} | vedclass

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(D) Let $S = 1 + \frac{1 	imes 3}{6} + \frac{1 	imes 3 	imes 5}{6 	imes 8} + \dots \infty$. Multiplying by $\frac{1}{4}$,we get $\frac{S}{4} = \fr

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$4$

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(D) Let $S = 1 + \frac{1 	imes 3}{6} + \frac{1 	imes 3 	imes 5}{6 	imes 8} + \dots \infty$. Multiplying by $\frac{1}{4}$,we get $\frac{S}{4} = \frac{1}{4} + \frac{1 	imes 3}{4 	imes 6} + \frac{1 	imes 3 	imes 5}{4 	imes 6 	imes 8} + \dots \infty$. Using the binomial expansion $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$,we compare the series. We know that $(1-x)^{-1/2} = 1 + \frac{1}{2}x + \frac{1 	imes 3}{2 	imes 4}x^2 + \dots$. By manipulating the given series into the form of a binomial expansion with $x = -1/2$ and $n = -1/2$,we find that the sum converges to $S = 4$.

The sum of the series $1 + \frac{1 \times 3}{6} + \frac{1 \times 3 \times 5}{6 \times 8} + \dots \infty$ is

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