If `(1+x)^n=C_(0)C_1c+C_(2)x^2+…..+C_(n)x^n` then show that the sum of the products of the `C_(i)` taken two at a time represented by :`Sigma_(0 le I lt) Sigma_( j le n) C_(i)C_(j)` "is equal to " 2^(2n-1)-(2n!)/(2.n! n !)`

Explanation

Since `(c_0+C_1+C_2 +……+C_(n-1)+C_(n))^2`
`=C_(0)^(2)+C_(1)^(2)+C_(2)^(2)+.......C_(n-1)^(2)+C_(n)^2+2(C_(0)C_(1)+C_(0)C_(2)+C_(0)C_(3)+....+C_(0)C_(n)+C_(1)C_(2)+C_1C_3+...C_1C_n+C_2C_3+C_2C_4+...+C_2C_n+...+C_(n-1)C_n)`
`(2^)2=""^(2n)C_(n)+2 underset(0 le i lt j le n )(SigmaSigma) C_(i)C_(j)`
Hence `underset(0 le i lt j le n )(SigmaSigma) C_(i)C_(j)=2""^(2n-1)-(2n!)/(2.n!n!)`