$\begin{aligned} & ext { If } \frac{x^3}{(2 x-1)(x-1)^2}=A+\frac{B}{2 x-1}+\frac{C}{x-1} \ & +\frac{D}{(x-1)^2}, ext { then } 2 A-3 B+4 C+5 D=\end{aligned}$

Question

$\begin{aligned} & 	ext { If } \frac{x^3}{(2 x-1)(x-1)^2}=A+\frac{B}{2 x-1}+\frac{C}{x-1} \ & +\frac{D}{(x-1)^2}, 	ext { then } 2 A-3 B+4 C+5 D=\end{aligned}$

Aicharya · Accepted Answer

We have, $\frac{x^3}{(2 x-1)(x-1)^2}=\frac{x^3}{2 x^3-5 x^2+4 x-1}$

Now, consider $ \begin{aligned} & \frac{1}{2}\left(\frac{5 x^2-4 x+1}{2 x^3-5 x^2+4 x-1}\right) \\ & \quad=\frac{1}{2}\left(\frac{B}{2 x-1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}\right) \\ & \frac{1}{2}\left(\frac{5 x^2-4 x+1}{2 x^3-5 x^2+4 x-1}\right) \\ & =\frac{1}{2}\left(\frac{B(x-1)^2+C(x-1)(2 x-1)+D(2 x-1)}{(2 x-1)(x-1)^2}\right) \\ & \Rightarrow 5 x^2-4 x+1 \\ & \quad=B(x-1)^2+C(x-1)(2 x-1)+D(2 x-1) \end{aligned} $ Put $x=1$, we get $ D=2 $ Put $x=1 / 2$, we get $B=+1$ Now put $x=0$, we get $ \begin{aligned} 1 & =B+C-D \\ \Rightarrow \quad 1 & =+1+C-2 \quad \Rightarrow \quad C=2 \end{aligned} $ Now, $\frac{1}{2}\left(\frac{5 x^2-4 x+1}{(2 x-1)(x-1)^2}\right)$ $ \begin{gathered} =\frac{1}{2}\left(\frac{+1}{2 x-1}+\frac{2}{x-1}+\frac{2}{(x-1)^2}\right) \\ =\frac{+\frac{1}{2}}{2 x-1}+\frac{\frac{2}{2}}{x-1}+\frac{1}{(x-1)^2} \end{gathered} $ Put in Eq. (i), we get $ =\frac{1}{2}+\left(\frac{\frac{1}{2}}{2 x-1}+\frac{1}{x-1}+\frac{1}{(x-1)^2}\right) $ By comparing $A=1 / 2, B=1 / 2, C=1, D=1$. Now, $ \begin{aligned} & 2 A-3 B+4 C+5 D \\ & =2 \times \frac{1}{2}-3 \times \frac{1}{2}+4 \times 1+5 \times 1 \\ & =1-\frac{3}{2}+4+5=10-\frac{3}{2}=\frac{17}{2} \end{aligned} $

$\begin{aligned} & \text { If } \frac{x^3}{(2 x-1)(x-1)^2}=A+\frac{B}{2 x-1}+\frac{C}{x-1} \\ &…

Solution