1+4+12+32+80+192+……………+n×2的n-1次方,
问题描述:
1+4+12+32+80+192+……………+n×2的n-1次方,
答
123213
答
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