采用分组求和.
S=[2+2²+2³+…+2^n]-[1+2+3+…+n]
.=[2-2^(n+1)]/[1-2]-[n(n+1)]/2
.=2^(n+1)-2-(1/2)n(n+1)
则:
Sn=2^(n+1)-(1/2)n(n+1)-2