Question

Prove that, $n\times 2^{n-1}={}^n{C}_1+2\times {}^n{C}_2+.....n\times {}^n{C}_n.$

Answer 1

We have,

$(1+x{)}^n={}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+.....+{}^n{C}_n{x}^n$;

Now differentiating both sides w.r.t. to 'x' we get,

$n(1+x{)}^{n-1}={}^n{C}_1+2{}^n{C}_{2}x+...\dots +n{}^n{C}_n{x}^{n-1}$

putting $x=1$ both sides we get,

$n{2}^{n-1}={}^n{C}_1+2{}^n{C}_2+...\dots +n{}^n{C}_n$.