The sum of the series 1 + frac{1}{5} + frac{1 | vedclass

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(C) The given series is $S = 1 + \frac{1}{5} + \frac{1 	imes 3}{5 	imes 10} + \frac{1 	imes 3 	imes 5}{5 	imes 10 	imes 15} + \dots$ We compare

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$\sqrt{\frac{5}{3}}$

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(C) The given series is $S = 1 + \frac{1}{5} + \frac{1 	imes 3}{5 	imes 10} + \frac{1 	imes 3 	imes 5}{5 	imes 10 	imes 15} + \dots$ We compare this with the binomial expansion for any index: $(1 - x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \dots$ Rewriting the series: $S = 1 + \frac{1}{5} + \frac{1 	imes 3}{2! 	imes 5^2 	imes 2^2} + \dots$ Alternatively,using the form $(1-x)^{-n} = 1 + \frac{n}{1}x + \frac{n(n+1)}{1 	imes 2}x^2 + \dots$,we identify $nx = \frac{1}{5}$ and $\frac{n(n+1)}{2}x^2 = \frac{3}{50}$. Solving for $n$ and $x$,we get $n = -\frac{1}{2}$ and $x = -\frac{2}{5}$. Thus,$S = (1 - (-\frac{2}{5}))^{-(-1/2)} = (1 + \frac{2}{5})^{-1/2} = (\frac{7}{5})^{-1/2}$ is incorrect. Correct approach: The series is of the form $(1-x)^{-n} = 1 + \frac{n}{1}x + \frac{n(n+1)}{2!}x^2 + \dots$ Here,$nx = \frac{1}{5}$ and $\frac{n(n+1)}{2}x^2 = \frac{3}{50}$. Dividing the second by the square of the first: $\frac{n(n+1)}{2n^2} = \frac{3/50}{1/25} = \frac{3}{2} \implies \frac{n+1}{n} = 3 \implies n+1 = 3n \implies n = \frac{1}{2}$. Then $x = \frac{1}{5n} = \frac{1}{5(1/2)} = \frac{2}{5}$. Since the signs in the series are all positive,we use $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \dots$ with $x$ replaced by $-x$,so $S = (1 - x)^{-n} = (1 - \frac{2}{5})^{-1/2} = (\frac{3}{5})^{-1/2} = \sqrt{\frac{5}{3}}$.

The sum of the series $1 + \frac{1}{5} + \frac{1 \times 3}{5 \times 10} + \frac{1 \times 3 \times 5}{5 \times 10 \times 15} + \dots$ is equal to

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