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Question

Prove that, (2nC1)2+2.(2nC1)2+3.(2nC3)2+....+2n.(2nC2n)2=(4n1)!{(2n1)!}2.

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Solution

(1+x)2n=2nC0+2nC1x+2nC2x2+....+2nC2nx2n(1)
Differentiating eq (1) w.r. to x

2n(1+x)2n1=2nC1+22nC2x+....+2n2nC2nx2n1(2)

We know that
(1+1x)2n=2nC0+2nC1x+2nC2x2+....+2nC2nx2n(3)

Multiplying eq (2) and (3) and comparing xr we get
2nC12nCr+1+22nC22nCr+2+......+2n2nC2n2nrC2n=2n4n1C2n+r

Putting r=0
(2nC1)2+2(2nC2)2+......+2n(2nC2n)2C2n=2n4n1C2n

(2nC1)2+2(2nC2)2+......+2n(2nC2n)2C2n=2n(4n1)!(2n)!(4n12n)!

(2nC1)2+2(2nC2)2+......+2n(2nC2n)2C2n=(4n1)!(2n1)!(2n1)!

(2nC1)2+2(2nC2)2+......+2n(2nC2n)2C2n=(4n1)!((2n1)!)2

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