Question

If $C_0,C_1,C_2,...C_n$ are the binomial coefficients in the expansion of $(1+x{)}^n$ then prove that:
$\frac{C_1}{C_0}+2\frac{C_2}{C_1}+3\frac{C_3}{C_2}+....+n\frac{C_n}{C_{n-1}}=\frac{n(n+1)}{2}$

Answer 1

$\frac{r^n{c}_r}{{}^n{c}_r-1}=\frac{rn!}{r!(n-r)!}\times \frac{(r-1)!}{(}n-r+1)!n!$ ${}^n{c}_r=\frac{n!}{r!(n-r)!}$

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

= n - r + 1

put r = 1 $\frac{c_1}{c_0}=n$

r = 2$\frac{2c_2}{c_1}=n-1$

$r=3\frac{3c_3}{c_2}=n-2$

$r=n\frac{n{c}_n}{c_{n-1}}=1$

Adding , we get

$\frac{c_1}{c_0}+\frac{{}^2{c}_2}{c_1}+\frac{{}^3{c}_3}{c_2}+\frac{{}^n{c}_n}{c_{n-1}}=n+n-1+n-2+...+1$

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