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If $\frac{2 x^3+3 x^2+3 x+5}{\left(x^2+1\right)\left(x^2+2\right)}$ is expanded in terms of the powers of $x$, then the coefficient of $x^5$ is
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Verified Answer
The correct answer is:
$\frac{9}{8}$
$$
\begin{aligned}
& \text { }\left(2 x^3+3 x^2+3 x+5\right)\left(1+x^2\right)^{-1}\left(2+x^2\right)^{-1} \\
& =2^{-1}\left(2 x^3+3 x^2+3 x+5\right)\left(1+x^2\right)^{-1}\left(1+\frac{x^2}{2}\right)^{-1} \\
& =2^{-1}\left(2 x^3+3 x^2+3 x+5\right) \times \\
& \left(1+(-1) x^2+x^4-x^6+\ldots\right)\left(1-\frac{x^2}{2}+\frac{x^4}{4}-\frac{x^6}{8}+\ldots\right)
\end{aligned}
$$
For coeff. of $x^5$.
$$
\begin{aligned}
& 2^{-1}\left[2(-1)+2(1)\left(\frac{-1}{2}\right)+3\left(\frac{1}{4}+\frac{1}{2}+1\right)\right] \\
& =2^{-1}\left[-2-1+\frac{21}{4}\right]=\frac{1}{2} \times \frac{9}{4}=\frac{9}{8} .
\end{aligned}
$$
\begin{aligned}
& \text { }\left(2 x^3+3 x^2+3 x+5\right)\left(1+x^2\right)^{-1}\left(2+x^2\right)^{-1} \\
& =2^{-1}\left(2 x^3+3 x^2+3 x+5\right)\left(1+x^2\right)^{-1}\left(1+\frac{x^2}{2}\right)^{-1} \\
& =2^{-1}\left(2 x^3+3 x^2+3 x+5\right) \times \\
& \left(1+(-1) x^2+x^4-x^6+\ldots\right)\left(1-\frac{x^2}{2}+\frac{x^4}{4}-\frac{x^6}{8}+\ldots\right)
\end{aligned}
$$
For coeff. of $x^5$.
$$
\begin{aligned}
& 2^{-1}\left[2(-1)+2(1)\left(\frac{-1}{2}\right)+3\left(\frac{1}{4}+\frac{1}{2}+1\right)\right] \\
& =2^{-1}\left[-2-1+\frac{21}{4}\right]=\frac{1}{2} \times \frac{9}{4}=\frac{9}{8} .
\end{aligned}
$$
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