Question

Prove that $\frac{C_0}{2}+\frac{C_1}{3}+\frac{C_2}{4}......+\frac{C_n}{n+2}=\frac{1+n\cdot 2^{n+1}}{(n+1)(n+2)}$

Answer 1

$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+....$
$\frac{C_2}{4}$ is desired. We have a turn of $C_2{x}^2$
Multiply by x, it becomes$C_2{x}^3$ and on integrating with limits 0 to 1 it becomes $C_2{x}^2$
Multiply both sides by x
$x(1+x{)}^n=C_{0}x+C_1\cdot x^2+C_2{x}^3+C_3{x}^4+....$
Now integrate w.r.t x within limits 0 to 1
$x\cdot \frac{(1+x{)}^{n+1}}{n+1}-\int 1\cdot \frac{(1+x{)}^{n+1}}{n+1}dx$
$={[\frac{(1+x{)}^{n+1}}{n+1}-\frac{(1+x{)}^{n+2}}{(n+1)(n+2)}]}_0^1$
$={[C_0\frac{x^2}{2}+C_1\frac{x^3}{3}+C_2\frac{x^4}{4}+...]}_0^1$
$\frac{2n^{n+1}}{n+1}-\frac{2n^{n+1}}{(n+1)(n+2)}-[0-\frac{n}{(n+1)(n+2)}]$
$=\frac{C_0}{2}+\frac{C_1}{3}+\frac{C_2}{4}+\frac{C_3}{5}+...+\frac{C_n}{n+2}$
$L.H.S=\frac{2^{n+1}(n+2-2)}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}$
= $\frac{n\cdot 2^{n+1}+1}{(n+1)(n+2)}$
This proves the result