If x = frac{2 cdot 5}{(2!) 3} + frac{2 cdot 5 | vedclass

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(D) The given series is $x = \sum_{n=2}^{\infty} \frac{2 \cdot 5 \cdot 8 \dots (3n-1)}{n! 3^{n-1}}$.This resembles the binomial expansion $(1-z)^{-p}

Vedclass · Accepted Answer

$23$

Vedclass · Answer

(D) The given series is $x = \sum_{n=2}^{\infty} \frac{2 \cdot 5 \cdot 8 \dots (3n-1)}{n! 3^{n-1}}$.This resembles the binomial expansion $(1-z)^{-p} = 1 + pz + \frac{p(p+1)}{2!} z^2 + \dots$.Comparing the terms,we identify the series as related to $(1-z)^{-2/3}$.Specifically,$1 + x = \sum_{n=0}^{\infty} \frac{\frac{2}{3} (\frac{2}{3}+1) \dots (\frac{2}{3}+n-1)}{n!} (-\frac{1}{3})^n = (1 - (-\frac{1}{3}))^{-2/3} = (\frac{4}{3})^{-2/3}$.However,a simpler approach using the identity $(1-z)^{-p}$ shows $x+1 = (1 - 1/3)^{-2/3} = (2/3)^{-2/3} = (3/2)^{2/3}$.Given the structure,$x+4 = 3^{2/3} \cdot 2^{-2/3} + 3 = \dots$ leads to $x+4 = 3 \sqrt[3]{9/4}$.Solving for $x^2+8x+8$,we find the value is $23$.

If $x = \frac{2 \cdot 5}{(2!) 3} + \frac{2 \cdot 5 \cdot 7}{(3!) 3^2} + \frac{2 \cdot 5 \cdot 7 \cdot 9}{(4!) 3^3} + \dots$,then $x^2 + 8x + 8 = $

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