求数列1+1/2，3+1/4，5+1/8，7+1/16……的前n项和

Question

求数列1+1/2，3+1/4，5+1/8，7/16……的前n项和

Answer 1

解：用拆项法
Sn=(1+1/2)+(3+1/4)+...+[(2n-1)+1/2^n]
=[1+3+...+(2n-1)]+(1/2+1/4+...+1/2^n)
=n[1+(2n-1)]/2 +1/2(1-1/2^n)/(1-1/2)
=n^2+1/2-1/2^n追问

那里来的(2n-1)+1/2^n]