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If $\frac{x+1}{x^4(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x+2}$, then $B+D+E$ is equal to
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Verified Answer
The correct answer is:
$A+C$
Given,
$$
\begin{aligned}
& \frac{x+1}{x^4(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x+2} \\
& \Rightarrow \quad x+1=A x^3(x+2)+B x^2(x+2)+C x(x+2) \\
&+D(x+2)+E x^4
\end{aligned}
$$
At $x=-1$,
$$
\begin{aligned}
\Rightarrow & & -1+1 & =-A+B-C+D+E \\
\Rightarrow & & 0 & =B+D+E-A-C \\
& \therefore & B+D+E & =A+C
\end{aligned}
$$
$$
\begin{aligned}
& \frac{x+1}{x^4(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x+2} \\
& \Rightarrow \quad x+1=A x^3(x+2)+B x^2(x+2)+C x(x+2) \\
&+D(x+2)+E x^4
\end{aligned}
$$
At $x=-1$,
$$
\begin{aligned}
\Rightarrow & & -1+1 & =-A+B-C+D+E \\
\Rightarrow & & 0 & =B+D+E-A-C \\
& \therefore & B+D+E & =A+C
\end{aligned}
$$
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