f(x)=1+x+x^2+x^3+...+x^n=(1-x^(n+1))/(1-x)
f'(x)=1+2x+3x^2+4x^3+...nx^(n-1)=[(1-x^(n+1))/(1-x)]'=((n (x-1)-1) x^n+1)/(x-1)^2
f'(2)=(n-1)*2^n+1
1+4+12+32+80+192+……………+n×2的n－1次方,
1+4+12+32+80+192+……………+n×2的n－1次方,
数学人气：118 ℃时间：2019-12-24 23:34:46
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