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

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

For any quadratic polynomial $f(x)$, it is true that $f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2!}(x-a)^2$ where $a$ is any real number. If $\frac{3 x^2+4 x+7}{(x-2)^3}=\frac{A}{(x-2)^3}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)}$ and $g(x)=3 x^2+4 x+7$, then $A+B+C=$

If $\frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{x-1}$, then $A+B+C+D+E=$

If $\frac{2x}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}$,then:

If $\frac{x+1}{(x-1)^2(x^2+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}$,then $\sqrt{3A^2+4D^2+5C^2+B^2}=$