Question

Prove that
$\frac{C_1}{C_0}+\frac{2C_2}{C_1}+\frac{3C_3}{C_2}+....+\frac{n.C_n}{C_{n-1}}=\frac{n(n+1)}{2}$

Answer 1

We have,

$\frac{k{C}_k}{C_{k-1}}=\frac{k(\frac{n}{k})}{(\frac{n}{k-1})}$

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

$=k\cdot \frac{(n-k+1)}{k}$

$=n-k+1$

$k=1\to \frac{C_1}{C_0}=n$

$k=2\to \frac{2C_2}{C_1}=n-1$

$k=3\to \frac{3C_3}{C_2}=n-2$

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$k=n\to \frac{n{C}_n}{C_{n-1}}=1$

On adding all the above terms, we get,

$\frac{C_1}{C_0}+\frac{2C_2}{C_1}+\frac{3C_3}{C_2}+.........+\frac{n{C}_n}{C_{n-1}}=n+(n-1)+(n-2)+......+1$

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

Hence proved.