an=Sn-Sn-1=(100n-n2)-［100·(n-1)-(n-1)2］=101-2n(n≥2).

∵a1=S1=100×1-12=99=101-2×1,

∴數(shù)列{an}的通項(xiàng)公式為an=101-2n(n∈N*).

又an+1-an=-2為常數(shù),

∴數(shù)列{an}是首項(xiàng)為a1=99,公差d=-2的等差數(shù)令an=101-2n≥0,得n≤50.5.

∵n∈N*,∴n≤50(n∈N*).

①當(dāng)1≤n≤50時(shí),an>0,此時(shí)bn=|an|=an,所以{bn}的前n項(xiàng)和Sn′=100n-n2.

②當(dāng)n≥51時(shí),an<0,此時(shí)bn=|an|=-an,

由b51+b52+…+bn=-(a51+a52+…+an)=-(Sn-S50)=S50-Sn,

    得數(shù)列{bn}的前n項(xiàng)和為

Sn′=S50+(S50-Sn)=2S50-Sn=2×2 

500-(100n-n2)=5 000-100n+n2.

    由①②得數(shù)列{bn}的前n項(xiàng)和為Sn′=