Question

$\frac{c_1}{c_0}+2\frac{c_2}{c_1}+3\frac{c_3}{c_2}+.........+n\frac{c_n}{c_{n-1}}=\frac{n(n+1)}{2}$

Answer 1

Consider the given function:

$\frac{c_1}{c_0}+\frac{2c_2}{c_1}+\frac{3c_3}{c_2}+\frac{4c_4}{c_3}+......+\frac{n{c}_n}{c_{n-1}}$

$=\frac{n}{1}+\frac{2n(n-1)}{2!}.\frac{1}{n}+\frac{3n(n-1)(n-2)}{3!}.\frac{1}{\frac{n(n-1)}{2!}}+......+\frac{n.1}{n}$

$=n+(n-1)+(n-2)+......1$

$=1+2+3+......+n$

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

Hence this is the answer.