If Cr stands for nCr then the sum of the series 2(n2)!(n2)n![C20−2C21+3C22−⋯+(−1)n(n+1)C2n], where n is an even positive integer is equal to
A
(−1)n2(n+2)
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B
(−1)n(n+2)
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C
(−1)nn
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Solution
The correct option is A(−1)n2(n+2) Since n is even, let n= 2m then, L.H.S. = S = 2×m!m!(2m)![C20−2C21+3C22+⋯+(−1)2m(2m+1)C22m]…(1) or S= 2×m!m!(2m)![(2m+1)C20−2mC21+(2m−1)C22+⋯+c22m]…(2) Adding (1) and (2), we get 2S=2m!m!(2m)!(2m+2)[C20−C21+C22+⋯+C22m] We know that C20−c21−⋯+C2n=(−1)n2nCn2S=2m!m!(2m)!(m+1)[(−1)m2mCm]=2(n2+1)(−1)n2=(−1)n2(n+2)