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Prove that ^nC0 "^(2n)Cn-^nC1 ^(2n-2)Cn ...

Prove that `^nC_0 "^(2n)C_n-^nC_1 ^(2n-2)C_n +^nC_2 ^(2n-4)C_n =2^n`

Text Solution

Verified by Experts

We know that
`.^(2n)C_(n) =` Coefficient of `x^(n)` in `(1+x)^(2n)`
`.^(2n-1)C_(n) =` Coefficient of `x^(n)` in `(1+x)^(2n-1)`
`.^(2n-2)C_(n) =` Coefficient of `x^(n)` in `(1+x)^(2n-2)`
`.^(2n-3)C_(n) =` Coefficient of `x^(n)` in `(1+x)^(2n-3)`
`.^(n)C_(n) =` Coefficient of `x^(n)` in `(1+x)^(n)`
Thus, we have
`.^(n)C_(0).^(2n)C_(n)-.^(n)C_(1).^(2n-1)C_(n)+.^(n)C_(2).^(2n-2)C_(n)+"...."+(-1)^(n).^(n)C_(n).^(n)C_(n)`
= Coefficient of `x^(n)` in `[C_(0)(1+x)^(2n)-C_(1)(1+x)^(2n-1)+C_(2)(1+x)^(2n-2)-C_(3)(1+x)^(2n-3)+"...."+(-1)^(n)C_(n)(1+x)^(n)]`
`=` Coefficient of `x^(m)` in `(1+x)^(n)[C_(0)(1+x)^(n)-C_(1)(1+x)^(n-1)+C_(2)(1+x)^(n-2)-C_(3)(1+x)^(n-3)+"......"+(-1)^(n)C_(n)]`
`=` Coefficient of `x^(n)` in `(1+x)^(n)[(1+x)-1]^(n)`
`=` Coefficient of `x^(n)` in `(1+x)^(n)x^(n)`
`= 1`
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