1)f'(x)=cosx
它的正根an=π/2+(n-1)π=(n-1/2)π,n=1,2,3,4,.
2)记bn=2^nan=π(n-1/2)2^n
Sn= π[0.5*2+1.5*2^2+...+(n-1/2)2^n]
2Sn=π[ 0.5*2^2+.+(n-3/2)2^n+(n-1/2)2^(n+1)]
两式相减：-Sn=π[0.5*2+2^2+2^3+..+2^n-(n-1/2)2^(n+1)]
-Sn=π[2^(n+1)-3-(n-1/2)2^(n+1)]
-Sn=π[(3/2-n)2^(n+1)-3]
得：Sn=π[(n-3/2)2^(n+1)+3]