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

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(B) Given the series $\alpha = \frac{5}{2! 3} + \frac{5 \cdot 7}{3! 3^2} + \frac{5 \cdot 7 \cdot 9}{4! 3^3} + \ldots$.We use the binomial expansion $(

Vedclass · Accepted Answer

$23$

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(B) Given the series $\alpha = \frac{5}{2! 3} + \frac{5 \cdot 7}{3! 3^2} + \frac{5 \cdot 7 \cdot 9}{4! 3^3} + \ldots$.We use the binomial expansion $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \ldots$.Let $n = -5/2$ and $x = -1/3$. Then $(1 - (-1/3))^{-5/2} = (4/3)^{-5/2} = (3/4)^{5/2} = \frac{3^{5/2}}{32}$.Alternatively,consider $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \ldots$.Comparing terms,we find $\alpha = (1 - 1/3)^{-5/2} - 1 - (5/2)(1/3) = (2/3)^{-5/2} - 1 - 5/6 = (3/2)^{5/2} - 11/6$.Actually,the series is $\alpha = \sum_{k=2}^{\infty} \frac{(2k+1)!!}{k! 3^{k-1} 2^{k-1}}$.Using the expansion $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \dots$,for $n=5/2$ and $x=1/3$,we get $(1-1/3)^{-5/2} = 1 + (5/2)(1/3) + \frac{(5/2)(7/2)}{2!}(1/3)^2 + \dots = 1 + 5/6 + \frac{35}{8 \cdot 9} + \dots = 1 + 5/6 + \frac{35}{72} + \dots$.The given series is $\alpha = (1-1/3)^{-5/2} - 1 - 5/6 = (2/3)^{-5/2} - 11/6 = (3/2)^{5/2} - 11/6$.Evaluating $\alpha^2 + 4\alpha = (\alpha+2)^2 - 4$.With $\alpha = 3^{3/2} - 2$,we have $\alpha + 2 = 3^{3/2} = 3\sqrt{3}$.Then $(\alpha+2)^2 - 4 = (3\sqrt{3})^2 - 4 = 27 - 4 = 23$.

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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