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Question: If \(C_{r}\) stands for \(n ⥂ C_{r}\), the sum of given series \(\frac{2(n/2)!(n/2)!}{n!}\lbrack C_...

If CrC_{r} stands for n⥂Crn ⥂ C_{r}, the sum of given series

2(n/2)!(n/2)!n![C02−2C12+3C22−......+(−1)n(n+1)Cn2]\frac{2(n/2)!(n/2)!}{n!}\lbrack C_{0}^{2} - 2C_{1}^{2} + 3C_{2}^{2} - ...... + ( - 1)^{n}(n + 1)C_{n}^{2}\rbrack where n is an even positive integer, is

A

0

B

(−1)n/2(n+1)( - 1)^{n/2}(n + 1)

C

(−1)n(n+2)( - 1)^{n}(n + 2)

D

(−1)n/2(n+2)( - 1)^{n/2}(n + 2)

Answer

(−1)n/2(n+2)( - 1)^{n/2}(n + 2)

Explanation

Solution

We have C02−2C12+3C22−........+(−1)n(n+1)Cn2C_{0}^{2} - 2C_{1}^{2} + 3C_{2}^{2} - ........ + ( - 1)^{n}(n + 1)C_{n}^{2}

= [C02−C12+C22−.......+(−1)nCn2]−[C12−2C22+3C32......+(−1)nn.Cn2]\lbrack C_{0}^{2} - C_{1}^{2} + C_{2}^{2} - ....... + ( - 1)^{n}C_{n}^{2}\rbrack - \lbrack C_{1}^{2} - 2C_{2}^{2} + 3C_{3}^{2}...... + ( - 1)^{n}n.C_{n}^{2}\rbrack = (−1)n/2.n⥂Cn/2−(−1)n/2−1.12n.nCn/2( - 1)^{n/2}.^{n} ⥂ C_{n/2} - ( - 1)^{n/2 - 1}.\frac{1}{2}n.^{n}C_{n/2} =(−1)n/2[1+n2]nCn/2( - 1)^{n/2}\left\lbrack 1 + \frac{n}{2} \right\rbrack^{n}C_{n/2}

Therefore the value of given expression

=2.n2!n2!n![(−1)n/2.(1+n2)n!n2!n2!]=(−1)n/2(n+2)= \frac{2.\frac{n}{2}!\frac{n}{2}!}{n!}\left\lbrack ( - 1)^{n/2}.\left( 1 + \frac{n}{2} \right)\frac{n!}{\frac{n}{2}!\frac{n}{2}!} \right\rbrack = ( - 1)^{n/2}(n + 2)