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If $\mathrm{x}=\frac{2.5}{2 ! 3}+\frac{2.5 .7}{3 ! 3^2}+\frac{2.5 \cdot 7 \cdot 9}{4 ! 3^3}+\ldots$, then $x^2+8 x+8=$
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The correct answer is:
100
$\because(1-x)^{-3 / 2}=1+\frac{3}{2} x$
$\begin{aligned} & +\frac{\frac{3}{2}\left(\frac{3}{2}+1\right) x^2}{2 !}+\frac{\frac{3}{2}\left(\frac{3}{2}+1\right)\left(\frac{3}{2}+2\right)}{3 !} x^3 \\ & +\frac{\frac{3}{2}\left(\frac{3}{2}+1\right)\left(\frac{3}{2}+2\right)\left(\frac{3}{2}+3\right)}{4 !} x^4+\ldots\end{aligned}$
putting $x=\frac{2}{3}$, we get
$\begin{aligned} & \left(1-\frac{2}{3}\right)^{-\frac{3}{2}}=2+\frac{5}{2 ! 3}+\frac{5.7}{3 ! 3^2}+\frac{5.7 .9}{4 ! 3^2}+\ldots \\ & \left(\frac{1}{3}\right)^{-\frac{3}{2}}=\frac{1}{2}\left[4+\frac{2.5}{2 ! 3}+\frac{2.5 .7}{3 ! 3^2}+\frac{2.5 .7 .9}{4 ! 3^2}+\ldots\right]\end{aligned}$
$\left[(3)^{\frac{3}{2}}\right]^2=\left[\frac{1}{2}(4+x)\right]^2 \Rightarrow x^2+8 x+8=100$
$\begin{aligned} & +\frac{\frac{3}{2}\left(\frac{3}{2}+1\right) x^2}{2 !}+\frac{\frac{3}{2}\left(\frac{3}{2}+1\right)\left(\frac{3}{2}+2\right)}{3 !} x^3 \\ & +\frac{\frac{3}{2}\left(\frac{3}{2}+1\right)\left(\frac{3}{2}+2\right)\left(\frac{3}{2}+3\right)}{4 !} x^4+\ldots\end{aligned}$
putting $x=\frac{2}{3}$, we get
$\begin{aligned} & \left(1-\frac{2}{3}\right)^{-\frac{3}{2}}=2+\frac{5}{2 ! 3}+\frac{5.7}{3 ! 3^2}+\frac{5.7 .9}{4 ! 3^2}+\ldots \\ & \left(\frac{1}{3}\right)^{-\frac{3}{2}}=\frac{1}{2}\left[4+\frac{2.5}{2 ! 3}+\frac{2.5 .7}{3 ! 3^2}+\frac{2.5 .7 .9}{4 ! 3^2}+\ldots\right]\end{aligned}$
$\left[(3)^{\frac{3}{2}}\right]^2=\left[\frac{1}{2}(4+x)\right]^2 \Rightarrow x^2+8 x+8=100$
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