Question

sn=n*（n+1)/2 +2*2+3*2^2+……+（n+1)*2^n 用错位相减法的过程

Answer 1

sn=n*（n+1)/2 +2*2+3*2^2+……+（n+1)*2^n

Sn-1=n(n-1)/2+2*2+3*2^2+……+n*2^(n-1)

an=Sn-Sn-1

= n(n+1)/2 -n(n-1)/2+(n+1)*2^n

=n+(n+1)*2^n