Question

Prove that $\frac{C_0}{1}-\frac{C_1}{4}+\frac{C_2}{9}-...+\frac{(-1{)}^n{C}_n}{n^2+2n+1}$
$=\frac{1}{n+1}+\frac{1}{2n+2}+...+\frac{1}{n^2+2n+1}.$

Answer 1

L.H.S. $=\frac{C_0}{1^2}-\frac{C_1}{2^2}+\frac{C_2}{3^2}-...(-1{)}^n\frac{C_n}{(n+1{)}^2}$
Since the term are +, -we consider the expansion of
$(1-x{)}^n=C_0-C_{1}x+C_2{x}^2-...+(-1{)}^n{C}_n{x}^n$
Integrating we get
$\frac{-(1-x{)}^{n+1}}{}+k=C_{0}x-C_1\frac{x^2}{2}+C_2\frac{x^3}{3}-...x=0$ gives $k=\frac{1}{n+1}$
$\therefore \frac{1}{n+1}[1-(1-x{)}^{n+1}]=C_{0}x-C_1\frac{x^2}{2}+C_2\frac{x^3}{3}-...$
We have $C_2\frac{x^3}{3}$ and we want $\frac{C_2}{3^2}$. Hence we divide by x and integrate again within limits 0 to 1
$\frac{1}{n+1}\left[\frac{1-(1-x{)}^{n+1}}{1-(1-x)}\right]=C_0-C_1\frac{x}{2}+C_2\frac{x^2}{3}-...$
L.H.S. is a G.P. whose r = 1 - x
$\therefore \frac{1}{n+1}[1+(1-x)+(1-x{)}^2+...]$
$=C_0-C_1\frac{x}{2}+C_2\frac{x^2}{3}-...$
Now integrate both sides within limits 0 to 1 and you get the desired result.