1+frac{4}{15}+frac{4 times 10}{15 times 30}+f | vedclass

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(B) The given series is $S = 1 + \frac{4}{15} + \frac{4 	imes 10}{15 	imes 30} + \frac{4 	imes 10 	imes 16}{15 	imes 30 	imes 45} + \ldots \inft

Vedclass · Accepted Answer

$\left(\frac{5}{3}\right)^{2 / 3}$

Vedclass · Answer

(B) The given series is $S = 1 + \frac{4}{15} + \frac{4 	imes 10}{15 	imes 30} + \frac{4 	imes 10 	imes 16}{15 	imes 30 	imes 45} + \ldots \infty$. This is of the form $1 + nx + \frac{n(n+1)}{2!} x^2 + \ldots = (1-x)^{-n}$. We can rewrite the terms as: $S = 1 + \frac{4}{15} + \frac{4 	imes 10}{15 	imes 30} + \frac{4 	imes 10 	imes 16}{15 	imes 30 	imes 45} + \ldots$ $= 1 + \frac{4}{15} + \frac{4 	imes 10}{15 	imes 30} + \frac{4 	imes 10 	imes 16}{15 	imes 30 	imes 45} + \ldots$ $= 1 + \frac{4}{15} + \frac{4 	imes 10}{2! 	imes 15^2} 	imes \frac{1}{2} + \ldots$ Comparing with $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!} x^2 + \ldots$,we have $nx = \frac{4}{15}$ and $\frac{n(n+1)}{2} x^2 = \frac{40}{450} = \frac{4}{45}$. Dividing the two: $\frac{n(n+1)x^2 / 2}{nx} = \frac{4/45}{4/15} \implies \frac{(n+1)x}{2} = \frac{1}{3} \implies (n+1)x = \frac{2}{3}$. Since $nx = \frac{4}{15}$,we have $nx + x = \frac{2}{3} \implies \frac{4}{15} + x = \frac{10}{15} \implies x = \frac{6}{15} = \frac{2}{5}$. Then $n(2/5) = 4/15 \implies n = 2/3$. Thus,$S = (1 - 2/5)^{-2/3} = (3/5)^{-2/3} = (5/3)^{2/3}$.

$1+\frac{4}{15}+\frac{4 \times 10}{15 \times 30}+\frac{4 \times 10 \times 16}{15 \times 30 \times 45}+\ldots \quad \infty=$

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