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Evaluate.^(n)C(0).^(n)C(2)+2.^(n)C(1).^(...

Evaluate`.^(n)C_(0).^(n)C_(2)+2.^(n)C_(1).^(n)C_(3)+3.^(n)C_(2).^(n)C_(4)+"...."+(n-1).^(n)C_(n-2).^(n)C_(n)`.

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The correct Answer is:
`n..^(2n-1)C_(n-3)+.^(2n)C_(n-2)`
`S = .^(n)C_(0).^(n)C_(2)+2.^(n)C_(1).^(n)C_(3)+3.^(n)C_(2).^(n)C_(4)+"....."+(n-1).^(n)C_(n-2).^(n)C_(n)`
` = .^(n)C_(0).^(n)C_(n-2)+2.^(n)C_(1).^(n)C_(n-3)+3.^(n)C_(2).^(n)C_(n-4)+"....."(n-1).^(n)C_(n-2).^(n)C_(0)`
`= underset(r=1)overset(n)sumr.^(n)C_(r-1).^(n)C_(n-r-1)`
`= underset(r=1)overset(n)sum((r-1)+1).^(n)C_(r-1).^(n)C_(n-r-1)`
`=underset(r=1)overset(n)sum[(r-1).^(n)C_(r-1).^(n)C_(n-r-1)+.^(n)C_(r-1).^(n)C_(n-r-1)]`
`= underset(r=1)overset(n)sum[n^(n-1)C_(r-2).^(n)C_(n-r-1)+.^(n)C_(r1).^(n)C_(n-r-1)]`
`= n^(2n-1)C_(n-3)+.^(2n)C_(n-2)`
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