Find the value of `(sumsum)_(0leiltjlen) (1+j)(""^(n)C_(i)+""^(n)C_(j))`.

Explanation

Here sum does nto change if we replace `i` by `n-1` and j by `n-j`.
By doing so, in fact we are writing the series in the reverse order.
Therefore, `S = underset(0leiltjlen)(sumsum)(i+j)(.^(n)C_(i)+.^(n)C_(j))" "(1)`
`=underset(0leiltjlen)(sumsum)(n-i+n-j)(.^(n)C_(n-i)+.^(n)C_(n-j))`
`=underset(0leiltjlen)(sumsum)(2n(-i+j))(.^(n)C_(i) + .^(n)C_(j))" "(2)`
Adding (1) and (2), we have
`2S = 2n underset(0leiltjlen)(sumsum)(.^(n)C_(i) +.^(n)C_(j))`
or `S = n xx n2^(n) = n^(2)2^(n)`