偶数项有m项
a2=3
a2m=4m-1
所以偶数项和=(3+4m-1)*m/2==2m²+m
奇数项也是m个
a1=1
a(2m-1)=2^(2m-1)+1
所以奇数项和=2^1+2^3+……+2^(2m-1)-1*m
=2*(1-4^m)/(1-4)-m
=(2/3)*4^m+2/3-m
所以S(2m)=2m²+m+(2/3)*4^m+2/3-m
=2m²+(2/3)*4^m+2/3