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Question
(2nC1)2+2×(2nC2)2+3.(2nC3)2.....+2n(2nC2n)2,When n = 100
(2nC1)2+2×(2nC2)2+3.(2nC3)2.....+2n(2nC2n)2,When n = 100
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Solution
The correct option is C 399!(199!)2
Each term of the expansion is of the form r×(2nCr)2
⇒sum=∑2nr=1r(2nCr)2
We know, 2nCr=2nr×2n−1Cr−1
⇒r⋅2nCr=2n⋅2n−1Cr−1
⇒sum=∑2nr=12n2n−1Cr−1×2nCr
=2n∑2nr=12n−1Cr−1×2nCr
=2n∑2nr=12n−1Cr−1×2nC2n−r
[∵nCr=nCn−r]
The sum of subscript is a constant =(r-1)+(2n-r)
=2n-1
∑2nr=12n−1Cr−1×2nC2n−r is the coefficient of x2n−1 in the expansion of (1+x)2n−1×(1+x)2n
To understand this, consider
(1+x)2n−1(1+x)2n=(2n−1C0+2n−1C1x+......2n−1C2n−1x2n−1)×(2nC0+2nC1x1+2nC2x2.......2nCnx2n)
399!(199!)2
Each term of the expansion is of the form r×(2nCr)2
⇒sum=∑2nr=1r(2nCr)2
We know, 2nCr=2nr×2n−1Cr−1
⇒r⋅2nCr=2n⋅2n−1Cr−1
⇒sum=∑2nr=12n2n−1Cr−1×2nCr
=2n∑2nr=12n−1Cr−1×2nCr
=2n∑2nr=12n−1Cr−1×2nC2n−r
[∵nCr=nCn−r]
The sum of subscript is a constant =(r-1)+(2n-r)
=2n-1
∑2nr=12n−1Cr−1×2nC2n−r is the coefficient of x2n−1 in the expansion of (1+x)2n−1×(1+x)2n
To understand this, consider
(1+x)2n−1(1+x)2n=(2n−1C0+2n−1C1x+......2n−1C2n−1x2n−1)×(2nC0+2nC1x1+2nC2x2.......2nCnx2n)
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