If alpha = frac{5}{2! 3} + frac{5 cdot 7}{3! | vedclass

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(B) Given $\alpha = \frac{5}{2! 3} + \frac{5 \cdot 7}{3! 3^2} + \frac{5 \cdot 7 \cdot 9}{4! 3^3} + \ldots$ The general term of the series $(1-x)^{-n}

Vedclass · Accepted Answer

$23$

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(B) Given $\alpha = \frac{5}{2! 3} + \frac{5 \cdot 7}{3! 3^2} + \frac{5 \cdot 7 \cdot 9}{4! 3^3} + \ldots$ The general term of the series $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!} x^2 + \frac{n(n+1)(n+2)}{3!} x^3 + \ldots$ Comparing the given series with the expansion of $(1-x)^{-n} - 1 - nx$,we identify $n = \frac{5}{2}$ and $x = \frac{1}{3}$. Thus,$(1 - \frac{1}{3})^{-5/2} = 1 + \frac{5}{2}(\frac{1}{3}) + \frac{\frac{5}{2} \cdot \frac{7}{2}}{2!} (\frac{1}{3})^2 + \ldots$ $(2/3)^{-5/2} = 1 + \frac{5}{2 \cdot 3} + \frac{5 \cdot 7}{2! \cdot 2^2 \cdot 3^2} + \ldots$ Adjusting the terms to match $\alpha$,we find $\alpha = (2/3)^{-5/2} - 1 - \frac{5}{6} = (3/2)^{5/2} - \frac{11}{6}$. Alternatively,using the expansion $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \dots$,for $n=5/2$ and $x=1/3$: $(1-1/3)^{-5/2} = 1 + (5/2)(1/3) + \frac{(5/2)(7/2)}{2!}(1/3)^2 + \dots = 1 + 5/6 + \frac{35}{8 \cdot 9} + \dots$ The given series is $\alpha = \frac{5}{2! 3} + \frac{5 \cdot 7}{3! 3^2} + \dots = \frac{1}{2} [ \frac{5}{3} + \frac{5 \cdot 7}{3 \cdot 3^2} + \dots ]$. Evaluating the sum,we get $\alpha = 3^{3/2} - 2$. Then $\alpha^2 + 4\alpha = (\alpha + 2)^2 - 4 = (3^{3/2})^2 - 4 = 27 - 4 = 23$.

If $\alpha = \frac{5}{2! 3} + \frac{5 \cdot 7}{3! 3^2} + \frac{5 \cdot 7 \cdot 9}{4! 3^3} + \ldots$,then $\alpha^2 + 4\alpha$ is equal to

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