已知数列an的通项公式an=（2n-1）+1/2的n次方,求Sn

分组求和Sn=a1+a2+a3+……+an=(1+1/2)+(3+1/4)+(5+1/8) +……+[(2n-1)+1/2^n]=(1+3+5+……+(2n-1))+( 1/2+1/4+1/8+……+1/2^n)=n(1+2n-1)/2+ 1/2(1-1/2^n)/(1-1/2)=n^2+1-1/2^n.