1+frac{1}{3}+frac{1 times 3}{3 times 6}+frac{ | vedclass

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(D) The given series is $S = 1 + \frac{1}{3} + \frac{1 	imes 3}{3 	imes 6} + \frac{1 	imes 3 	imes 5}{3 	imes 6 	imes 9} + \ldots \infty$. We ca

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$\sqrt{3}$

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(D) The given series is $S = 1 + \frac{1}{3} + \frac{1 	imes 3}{3 	imes 6} + \frac{1 	imes 3 	imes 5}{3 	imes 6 	imes 9} + \ldots \infty$. We can rewrite the general term by multiplying the denominator by $3^n n!$: $S = 1 + \frac{1}{3(1!)} + \frac{1 	imes 3}{3^2(2!)} + \frac{1 	imes 3 	imes 5}{3^3(3!)} + \ldots$. Comparing this with the binomial expansion $(1-x)^{-p/q} = 1 + \frac{p}{q}x + \frac{p(p+q)}{2!}(\frac{x}{q})^2 + \frac{p(p+q)(p+2q)}{3!}(\frac{x}{q})^3 + \ldots$. Here,$p=1, q=2$,and $\frac{x}{q} = \frac{1}{3} \Rightarrow x = \frac{2}{3}$. Thus,$S = (1 - \frac{2}{3})^{-1/2} = (\frac{1}{3})^{-1/2} = \sqrt{3}$.

$1+\frac{1}{3}+\frac{1 \times 3}{3 \times 6}+\frac{1 \times 3 \times 5}{3 \times 6 \times 9}+\ldots \text{ to } \infty =$

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