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Question

If Cr stands for nCr then the sum of the series
2(n2)!(n2)n![C202C21+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)![C202C21+3C22++(1)2m(2m+1)C22m](1)
or S= 2×m!m!(2m)![(2m+1)C202mC21+(2m1)C22++c22m](2)
Adding (1) and (2), we get
2S=2m!m!(2m)!(2m+2)[C20C21+C22++C22m]
We know that C20c21+C2n=(1)n2 nCn2S=2m!m!(2m)!(m+1)[(1)m 2mCm]=2(n2+1)(1)n2=(1)n2(n+2)

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