题目内容
设数列{an}的前n项和为Sn=10n-n2,则|a1|+|a2|+…+|a15|等于( )
A.150 | B.135 | C.125 | D.100 |
根据an=
,得
当n≥2时,an=Sn-Sn-1=-n2+10n-[-(n-1)2+10(n-1)]=-2n+11,
当n=1时,S1=a1=9也适合上式,
∴an=-2n+11,
据通项公式得a1>a2>…>a5>0>a6>a7>…>a15
∴|a1|+|a2|+…+|a15|
=(a1+a2+…+a5)-(a6+a7+…+a15)
=2S5-S15
=2×(10×5-52)-(10×15-152)
=50+75
=125.
故选C.
|
当n≥2时,an=Sn-Sn-1=-n2+10n-[-(n-1)2+10(n-1)]=-2n+11,
当n=1时,S1=a1=9也适合上式,
∴an=-2n+11,
据通项公式得a1>a2>…>a5>0>a6>a7>…>a15
∴|a1|+|a2|+…+|a15|
=(a1+a2+…+a5)-(a6+a7+…+a15)
=2S5-S15
=2×(10×5-52)-(10×15-152)
=50+75
=125.
故选C.
练习册系列答案
相关题目