问题
解答题
已知{an}为等差数列,且a2=-1,a5=8.
(1)求数列{|an|}的前n项和;
(2)求数列{2n·an}的前n项和.
答案
(1) Sn= (2) 20+(3n-10)×2n+1
(1)设等差数列{an}的公差为d,因为a2=-1,a5=8,所以解得a1=-4,d=3,所以an=-4+3(n-1)=3n-7,因此|an|=|3n-7|=
,记数列{|an|}的前n项和为Sn,
当n=1时,S1=|a1|=4,当n=2时,S2=|a1|+|a2|=5,
当n≥3时,Sn=S2+|a3|+|a4|+…+|an|=5+(3×3-7)+(3×4-7)+…+(3n-7)=5+=
n2-
n+10.
又当n=2时满足此式,
综上,Sn=
(2)记数列{2nan}的前n项和为Tn
则Tn=2a1+22a2+23a3+…+2nan,2Tn=22a1+23a2+24a3+…+2nan-1+2n+1an,
所以-Tn=2a1+d(22+23+…+2n)-2n+1an
由(1)知,a1=-4,d=3,an=3n-7,所以-Tn=-8+3×-
=-20-(3n-10)×2n+1,故Tn=20+(3n-10)×2n+1.