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Find the following sum: sumsum(i ne j)...

Find the following sum:
`sumsum_(i ne j) ""^(n)C_(i).""^(n)C_(j)`

Text Solution

Verified by Experts

(i) `underset(i ne j)(sumsum).^(n)C_(i)..^(n)C_(j)=(underset(i=0)overset(n)sumunderset(j=0)overset(n)sum.^(n)C_(i).^(n)C_(j))-underset(i=0)overset(n)sum(.^(n)C_(i))^(2)`
`= (underset(i=0)overset(n)sum.^(n)C_(i))(underset(j=0)overset(n)sum.^(n)C_(j))-underset(i=0)overset(n)sum(.^(n)C_(i))^(2)`
`= (2^(n))(2^(n))-.^(2n)C_(n) = 4^(n)-.^(2n)C_(n)`
(ii) `underset(0leiltjlen)(sumsum).^(n)C_(i)..^(n)C_(j)= ((underset(i=0)(sum)underset(j=0)(sum).^(n)C_(i).^(n)C_(j))-underset(i=0)(sum)(.^(n)C_(i))^(2))/(2)`
`= (2^(2n)-.^(2n)C_(n))/(2)`
(ii) `underset(0leiltjlen)(sumsum).^(n)C_(i)..^(n)C_(j)= ((underset(i=0)(sum)underset(j=0)(sum).^(n)C_(i).^(n)C_(j))+underset(i=0)overset(n)sum(.^(n)C_(i))^(2))/(2)`
`= (2^(2n) + .^(2n)C_(n))/(2)`
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