(1+x)2n=2nC0+2nC1x+2nC2x2+....+2nC2nx2n−−−−−(1)
Differentiating eq (1) w.r. to x
2n(1+x)2n−1=2nC1+22nC2x+....+2n⋅2nC2nx2n−1−−−−−(2)
We know that
(1+1x)2n=2nC0+2nC1x+2nC2x2+....+2nC2nx2n−−−−−(3)
Multiplying eq (2) and (3) and comparing xr we get
2nC12nCr+1+2⋅2nC22nCr+2+......+2n⋅2nC2n2n−rC2n=2n⋅4n−1C2n+r
Putting r=0
(2nC1)2+2⋅(2nC2)2+......+2n⋅(2nC2n)2C2n=2n⋅4n−1C2n
(2nC1)2+2⋅(2nC2)2+......+2n⋅(2nC2n)2C2n=2n⋅(4n−1)!(2n)!(4n−1−2n)!
(2nC1)2+2⋅(2nC2)2+......+2n⋅(2nC2n)2C2n=⋅(4n−1)!(2n−1)!(2n−1)!
(2nC1)2+2⋅(2nC2)2+......+2n⋅(2nC2n)2C2n=⋅(4n−1)!((2n−1)!)2