Question

For $r=0,1,2,....,n$, prove that $C_0\cdot C_r+C_1\cdot C_{r+1}+C_2\cdot C_{r+2}+....+C_{n-r}\cdot C_n={}^{2n}{C}_{(n+r)}$ and hence deduce that
i) $C_0^2+C_1^2+C_2^2+......+C_n^2={}^{2n}{C}_n$
ii) $C_0\cdot C_1+C_1\cdot C_2+C_2\cdot C_3+.....+C_{n-1}\cdot C_n={}^{2n}{C}_{n+1}$

Answer 1

To prove:

$C_0.C_r+C_1.C_{r+1}+C_2.C_{r+2}+......+C_{n-r}.C_n={}^{2n}{C}_{n+r}$

$i)C_0^2+C_1^2+C_2^2+........+C_n^2={}^{2n}{C}_n$

$ii)C_0.C_1+C_1.C_2+C_2.C_3+......+C_{n-1}.C_n={}^{2n}{C}_{n+1}$

Solution:

We know that,

$C_0+C_{1}x+C_2{x}^2+......+C_n{x}^n=(1+x{)}^n$ $...............\left(1\right)$

$C_0{x}^n+C_1{x}^{n-1}+C_2{x}^{n-2}+......+C_n=(x+1{)}^n$ $...............\left(2\right)$

Multiplying eqn.$\left(1\right)$ and eqn.$\left(2\right)$ we get,

$(C_0+C_{1}x+C_2{x}^2+......+C_n{x}^n)$ $(C_0{x}^n+C_1{x}^{n-1}+C_2{x}^{n-2}+......+C_n)=(1+x{)}^{2n}$ $.............\left(3\right)$

Equating coeffiecients of $x^{n-r}$ from both sides of $\left(3\right)$ we get,

$C_0.C_r+C_1.C_{r+1}+C_2.C_{r+2}+......+C_{n-r}.C_n={}^{2n}{C}_{n+r}$ $..........\left(4\right)$

Now putting $r=0$ in eqn.$\left(4\right)$ we get,

$C_0.C_0+C_1.C_1+C_2.C_2+......+C_n.C_n={}^{2n}{C}_n$

or, $C_0^2+C_1^2+C_2^2+........+C_n^2={}^{2n}{C}_n$

Now putting $r=1$ in eqn.$\left(4\right)$ we get,

$C_0.C_1+C_1.C_2+C_2.C_3+......+C_{n-1}.C_n={}^{2n}{C}_{n+1}$