If $\frac{6 x^3+7 x^2+6 x-3}{(x-1)(x+3)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B}{x+3}+\frac{C x+D}{x^2+1}$…
If $\frac{6 x^3+7 x^2+6 x-3}{(x-1)(x+3)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B}{x+3}+\frac{C x+D}{x^2+1}$
and $n=A+B+C+D$ and ${ }^{50} C_n={ }^{50} C_r$, then $r$ is equal to
$40$
$43$
$35$
$42$
Solution
Given, $\frac{6 x^3+7 x^2+6 x-3}{(x-1)(x+3)\left(x^2+1\right)}$
$=\frac{A}{x-1}+\frac{B}{x+3}+\frac{(C x+D)}{x^2+1}$
$\Rightarrow \quad 6 x^3+7 x^2+6 x-3$
$\begin{aligned} & =A(x+3)\left(x^2+1\right)+B(x-1)\left(x^2+1\right) \\ & +C x(x-1)(x+3)+D(x-1)(x+3)\end{aligned}$
Put $x=1, A=2$
Put $x=-3, B=3$
Put $x=0, D=2$
Put $x=-1, A=3, B=3, D=2$, we get
$C=1$
$\therefore \quad n=A+B+C+D=2+3+1+2=8$
${ }^{50} C_n={ }^{50} C_r$
$\therefore n+r=50$
$r=52-n=50-8=42$