A) 0
B) \[{{(-1)}^{n/2}}(n+1)\]
C) \[{{(-1)}^{n}}(n+2)\]
D) \[{{(-1)}^{n/2}}(n+2)\]
Correct Answer: D
Solution :
We have \[C_{0}^{2}-2C_{1}^{2}+3C_{2}^{2}-.....+{{(-1)}^{n}}(n+1)C_{n}^{2}\] \[=[C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-....+{{(-1)}^{n}}C_{n}^{2}]-\]\[[C_{1}^{2}-2C_{2}^{2}+3C_{3}^{2}....-{{(-1)}^{n}}nC_{n}^{2}]\] =\[{{(-1)}^{n/2}}\frac{n!}{(n/2)!(n/2)!}-{{(-1)}^{(n/2)-1}}.\frac{1}{2}n\,{{\,}^{n}}{{C}_{n/2}}\] =\[{{(-1)}^{n/2}}.\frac{n!}{(n/2)!(n/2)!}.\left( 1+\frac{n}{2} \right)\] Therefore the value of the given expression is \[\frac{2(n/2)\,!\,(n/2)!}{n!}\times {{(-1)}^{n/2}}.\frac{(n)!}{(n/2)!(n/2)!}\left( 1+\frac{n}{2} \right)\] \[={{(-1)}^{n/2}}(2+n)\]
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