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Question
Find the sum of the series nC0nC2+nC1nC3+nC2nC4.....nCn−2nCn
Find the sum of the series nC0nC2+nC1nC3+nC2nC4.....nCn−2nCn
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Solution
The correct option is C 2nCn−2
We can re-write the terms as nC0nCn−2+nC1nCn−3+nC2nCn−4.....nCn−2nC0
We wrote it this way because, we can see that the p numbers in the subscript add up to n - 2 (e.g. In nC0nCn−2, the subscripts are n - 2 and 0)
⇒ Lets consider the expansion of (1+x)n(x+n)n and find the coefficient of xn−2.
(1+x)n(x+1)n=(nC0+nC1x+nC2x2.....nCn−2xn−2+nCn−1xn−1+nCnxn) × (nC0xn+nC1xn−1.....nCn)
Coefficient of xn−2=nC0×nCn−2+nC1nCn−3+.....nCn−2×nC0
⇒ The required sum is the coefficient xn−2 in (1+x)n(x+1)n or (1+x)2n.
=2nCn−2
2nCn−2
We can re-write the terms as nC0nCn−2+nC1nCn−3+nC2nCn−4.....nCn−2nC0
We wrote it this way because, we can see that the p numbers in the subscript add up to n - 2 (e.g. In nC0nCn−2, the subscripts are n - 2 and 0)
⇒ Lets consider the expansion of (1+x)n(x+n)n and find the coefficient of xn−2.
(1+x)n(x+1)n=(nC0+nC1x+nC2x2.....nCn−2xn−2+nCn−1xn−1+nCnxn) × (nC0xn+nC1xn−1.....nCn)
Coefficient of xn−2=nC0×nCn−2+nC1nCn−3+.....nCn−2×nC0
⇒ The required sum is the coefficient xn−2 in (1+x)n(x+1)n or (1+x)2n.
=2nCn−2
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