duoduo1984
幼苗
共回答了12个问题采纳率:100% 举报
(Ⅰ)设等差数列{an}的公差为d,等比数列{bn}的公比为q,则
11+d+q=11
1+2d+q2=11,解得
q=2
d=−2,
所以an=-2n+13,bn=2n-1;
(Ⅱ)由(Ⅰ)得:|an-bn|=|13-2n-2n-1|;
(i)0<n≤3时,an>bn,an-bn=13-2n-2n-1,Sn=
(11−2n+13)n
2-
1×(1−2n)
1−2=-2n-n2+12n+1,且S3=20;
(ii)n>3时,an<bn,|an-bn|=bn-an=2n-1-(13-2n),
Sn-S3=
8(1−2n−3)
1−2-
(5−2n+13)(n−3)
2=2n+n2-12n+19,
∴Sn=2n+n2-12n+39;
综上所述,Sn=
−2n−n2+12n+1(0<n≤3)
2n+n2−12n+39(n≥4)
1年前
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