已知:数列{an}的通项公式为an=3n-1(n∈N*),等差数列{bn}中,bn>0且b1+b2+b3=15又a1+b1,a2+b2,a3+b

2025-05-14 21:13:10
推荐回答(1个)
回答1:

(1)由题意可得b1+b2+b3=3b2=15,即b2=5,
又由题意可得(a2+b22=(a1+b1)(a3+b3),
设等差数列{bn}的公差为d,
代入数据可得(3+5)2=(1+5-d)(9+5+d),
解之可得d=-10,或d=2,当d=-10不满足bn>0应舍去,
故d=2,bn=5+2(n-2)=2n+1;
(2)可得cn=

1
bn2?1
=
1
(2n+1)2?1
=
1
4n(n+1)
=
1
4
1
n
?
1
n+1
),
故数列{cn}的前n项和为:
1
4
(1-
1
2
+
1
2
?
1
3
+…+
1
n
?
1
n+1
)=
1
4
(1-
1
n+1
)=
n
4(n+1)