Question

Solve $\frac{C_1}{C_0}+2\frac{C_2}{C_1}+3\frac{C_3}{C_2}+\cdots \cdots +n\frac{C_n}{C_{n-1}}=\frac{n\left(n+1\right)}{2}$

Answer 1

$LHS=\frac{{}^n{C}_1}{{}^n{C}_0}+2\frac{{}^n{C}_2}{{}^n{C}_1}+3\frac{{}^n{C}_3}{{}^n{C}_2}+...+n\frac{{}^n{C}_n}{{}^n{C}_{n-1}}$

$=\sum _{r=1}^{n}r\frac{{}^n{C}_r}{{}^n{C}_{r-1}}$ .........$\left(1\right)$

Now,$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}}$

$=\frac{(r-1)!(n-r+1)!}{r!(n-r)!}$

$=\frac{(r-1)!(n-r+1)(n-r)!}{r(r-1)!(n-r)!}$

$=\frac{n-r+1}{r}$

Hence $\left(1\right)$ becomes,

$S=\sum _{r=1}^{n}r\left(\frac{n-r+1}{r}\right)$

$=\sum _{r=1}^n(n-r+1)$

$=\sum _{r=1}^{n}n-\sum _{r=1}^{n}r+\sum _{r=1}^{n}1$

$=n\sum _{r=1}^{n}1-\sum _{r=1}^{n}r+\sum _{r=1}^{n}1$

$=(n+1)\sum _{r=1}^{n}1-\sum _{r=1}^{n}r$

$=n(n+1)-[1+2+3+...+n]$

$=n(n+1)-\frac{n(n+1)}{2}$........ since sum of first $n$ natural numbers$=\frac{n(n+1)}{2}$

$=\frac{n(n+1)}{2}=RHS$

Hence proved.