已知数列an满足a1+2a2+2^2a3+...+2^n-1an=n/2

a1+2a2+2^2a3+...+2^n-1an=n/2 ①
a1+2a2+2^2a3+...+2^(n-2)a(n-1)=(n-1)/2 ②
①-②
2^(n-1)an=1/2
an=1/(2^n) (n>1)
a1=1/2满足an
∴an=1/(2^n)
bn=n/(2^n)
sn用一下错位相减
sn=2-(1+n/2)(1/2)^(n-1)