JEE Advance - Mathematics (1983 - No. 25)
If $${\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ..... + {C_n}{x^n}$$ then show that the sum of the products of the $${C_i}s$$ taken two at a time, represented $$\sum\limits_{0 \le i < j \le n} {\sum {{C_i}{C_j}} } $$ is equal to $${2^{2n - 1}} - {{\left( {2n} \right)!} \over {2{{\left( {n!} \right)}^2}}}$$
2^(2n-1) - (2n)! / (2 * (n!)^2)
2^(2n) - (2n)! / ((n!)^2)
2^(2n-1) + (2n)! / (2 * (n!)^2)
2^(2n) + (2n)! / ((n!)^2)
2^(n-1) - (n)! / (2 * (n!)^2)
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