Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+....+C_n{x}^n$ then show that the sum of the products of the $C_i^{\prime }s$ taken two at a time, represented by $\underset{0\le i<j\le n}{\sum \sum C_i{C}_j}$ is equal to $2^{2n-1}-\frac{2n!}{2(n!{)}^2}$.

Answer 1

We know:

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

Also,

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

Multiplying eq(1) and eq(2):

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

$\therefore$ Co-eff of $x^n$ on both sides:

${}^{2n}{C}_n=C_0^2+C_1^2+C_2^2+....+C_n^2$ ..... $\left(4\right)$

Putting $x=1$ in eq $\left(3\right)$

$\Rightarrow 2^{2n}=C_0(C_0+C_1+...+C_n)+C_1(C_0+C_1+...+C_n)+.....$

$\Rightarrow 2^{2n}=(C_0^2+C_1^2+C_2^2+....+C_n^2)+2\sum \sum _{0\le i<j\le n}{C}_i{C}_j$

$\Rightarrow 2^{2n}={}^{2n}{C}_n+2\sum \sum _{0\le i<j\le n}{C}_i{C}_j$ .............. [From eq $\left(4\right)$]

$\Rightarrow 2^{2n}-\frac{\left(2n\right)!}{n!n!}=2\sum \sum _{0\le i<j\le n}{C}_i{C}_j$

$\Rightarrow \sum \sum _{0\le i<j\le n}{C}_i{C}_j=2^{2n-1}-\frac{\left(2n\right)!}{2(n!{)}^2}$

(Hence Proved)