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If (1+x)^n=C(0)+C(1)x+C(2)x^2+….+C(n)x^n...

If `(1+x)^n=C_(0)+C_(1)x+C_(2)x^2+….+C_(n)x^n` then prove that `(SigmaSigma)_(0 le i lt j le n ) C_(i)C_(j)^2=(n-1)^(2n)C_(n)+2^(2n)`

Text Solution

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L.H.S = `underset(0 le i lt j le n )(SigmaSigma) C_(i)C_(j)^2=(n-1)^(2n)C_(n)+2^(2n)`
`(C_(0)+C_(1))^2+(C_(0)+C_2)^2+....+(C_(0)+C_(n))^2+(C_1C_2)^2+(C_1+C_3)^2+....+(C_1+C_c)^2+(C_2+C_3)^2+(C_(2)+C_4)^2+....+(C_2+C_n)^2+......+(C_(n-1)+C_n)^2`
`=n(C_(0)^2+C_1^2+C_2^2+....+C_n^2)+2 underset(0 le i lt j le n )(SigmaSigma) C_(i)C_(j)`
`n.""^(2n)C_(n)+2{2^(2n-1)-(2n !)/(2.n!n!)} " " {"from Illustration 17 "}`
`=n.""^(2n)C_n+2^(2n)-""^(2n)C_n= (n-1).""^(2n)C_(n)+2^(2n)= R.H.S`
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