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Question

Find the sum of the series nC0nC2+nC1nC3+nC2nC4.....nCn2nCn



Solution

The correct option is B


We can re-write the terms as nC0nCn2+nC1nCn3+nC2nCn4.....nCn2nC0
We wrote it this way because, we can see that the p numbers in the subscript add up to n - 2 (e.g. In nC0nCn2, the subscripts are n - 2 and 0)
Lets consider the expansion of (1+x)n(x+n)n and find the coefficient of xn2.
(1+x)n(x+1)n=(nC0+nC1x+nC2x2.....nCn2xn2+nCn1xn1+nCnxn) × (nC0xn+nC1xn1.....nCn)
Coefficient of xn2=nC0×nCn2+nC1nCn3+.....nCn2×nC0
The required sum is the coefficient xn2 in (1+x)n(x+1)n or (1+x)2n.
=2nCn2

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