Question

Prove that $\frac{1}{2}{C}_1-\frac{2}{3}{C}_2+\frac{3}{4}{C}_3-\frac{4}{5}{C}_4+......+(-1{)}^{n+1}\cdot \frac{n}{n+1}=\frac{1}{n+1}$

Answer 1

L.H.S = $(1-\frac{1}{2})C_1-(1-\frac{1}{3})C_2+(1-\frac{1}{4})C_3...+$
$(C_1-C_2+C_3-C_4+...)-(\frac{C_1}{2}-\frac{C_2}{3}+\frac{C_3}{4}-...)$
Now $(C_0-C_1+C_2-C_3+...=0)$
and $\frac{C_1}{2}-\frac{C_2}{3}+\frac{C_3}{4}....=C_0-\frac{1}{n+1}$ by 3 last page
$\therefore L.H.S=C_0-(C_0-\frac{1}{n+1})=\frac{1}{n+1}$