数列1,2+1/2,3+1/2+1/4,…,n+1/2+1/4+…+1/2n-1的前n项和为
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1,2+1/2,3+1/2+1/4,…,n+1/2+1/4+…+1/2n-1 最后一项应该是1/2^(n-1)吧.

sn=1+2+1/2+3+1/2+1/4+…+n+1/2+1/4+…1/2^(n-1)

=1+2+3+...+n+1/2+1/2+1/4+…+n+1/2+1/4+…1/2^(n-1)

=n(n+1)/2+(1-1/2)+(1-1/4)+...+[1-1/2^(n-1)]

=n(n+1)/2+1-1/2+1-1/4+...+1-1/2^(n-1)

=n(n+1)/2+(n-1)-[1/2+1/4+...+1/2^(n-1)]

=n(n+1)/2+(n-1)-[1-1/2^(n-1)]

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

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