问题
解答题
已知数列{an}的前n项和Sn=10n-n2,(n∈N*).
(1)求a1和an;
(2)记bn=|an|,求数列{bn}的前n项和.
答案
(1)∵Sn=10n-n2,∴a1=S1=10-1=9.------------------(2分)
当n≥2,n∈N*时,Sn-1=10(n-1)-(n-1)2=10n-n2+2n-11
∴an=Sn-Sn-1=(10n-n2)-(10n-n2+2n-11)=-2n+11-------------------(4分)
又n=1时,a1=-2×1+11=9,符合已知条件.
∴an=-2n+11(n∈N*)----------------(5分)
(2)∵an=-2n+11,∴bn=|an|=-2n+11(n≤5) 2n-11(n>5)
设数列{bn}的前n项和为Tn,n≤5时,Tn=
=10n-n2,-------------------(8分)n(9-2n+11) 2
n>5时Tn=T5+
=25+(n-5)(b6+bn) 2
=25+(n-5)2=n2-10n+50(n-5)(1+2n-11) 2
故数列{bn}的前n项和Tn=
---------------------(12分)10n-n2(n≤5) n2-10n+50(n>5)