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If (1+a)^(n)=.^(n)C(0)+.^(n)C(1)a+.^(n)C...

If `(1+a)^(n)=.^(n)C_(0)+.^(n)C_(1)a+.^(n)C_(2)a^(2)+ . . +.^(n)C_(n)a^(n)`, then prove that
`.^(n)C_(1)+2.^(n)C_(2)+3.^(n)3C_(3)+ . . .+n.^(n)C_(n)=n.2^(n-1)`.

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