一道数列题已知数列an为等差数列,bn为单调递增的等比数列,且a1+a2+a3=-27.b1b2b3=512,a1+b1=｜a2+b2｜=a3+b3.(1)求a2+b2的值及数列an,bn的通项(2)若cn=(bn)/[(bn-2)(bn-1)],求数列cn的前n项和sn

Question

一道数列题
已知数列an为等差数列,bn为单调递增的等比数列,且a1+a2+a3=-27.b1b2b3=512,a1+b1=｜a2+b2｜=a3+b3.
(1)求a2+b2的值及数列an,bn的通项
(2)若cn=(bn)/[(bn-2)(bn-1)],求数列cn的前n项和sn

Answer 1

1.
设{an}公差为d,{bn}公比为q.
a1+a2+a3=3a2=-27
a2=-9
b1b2b3=b2³=512
b2=8
数列{bn}是单调递增的等比数列,b3>b2,因此公比q>1.
a2+b2=-9+8=-1
|a2+b2|=1
a1+b1=1 (1)
a3+b3=1 (2)
(1)+(2)
(a1+a3)+(b1+b3)=2
2a2+(b2/q+ b2q)=2
a2=-9 b2=8代入,整理,得
2q²-5q+2=0
(2q-1)(q-2)=0
q=1/2(舍去)或q=2
a3=1-b3=1-b2q=1-2×2=-3
a3-a2=d=(-3)-(-9)=6
a1=a2-d=-9-6=-15
an=a1+(n-1)d=-15+6(n-1)=6n-21
bn=b2q^(n-2)=8×2^(n-2)=2^(n+1)
数列{an}的通项公式为an=6n-21；数列{bn}的通项公式为bn=2^(n+1)
(2)
cn=bn/[(bn -2)(bn -1)]
=2^(n+1)/[(2^(n+1) -2)(2^(n+1) -1]
=2^n/[(2^n -1)(2×2^n -1)]
=1/(2^n -1) -1/[2^(n+1) -1]
Sn=c1+c2+...+cn
=1/(2-1) -1/(2²-1)+1/(2²-1)-1/(2³-1)+...+1/(2^n -1)- 1/[2^(n+1) -1]
=1 -1/[2^(n+1) -1]