已知数列{an}是等差数列,且满足:a1+a2+a3=6,a5=5;数列{bn}满足:bn-bn-1=an-1(n≥2,n∈N﹡),b1=1. (Ⅰ)求an和bn; (Ⅱ)记数列cn=1/bn+2n(n∈N*),若{cn}的前n项和为Tn,
问题描述:
已知数列{an}是等差数列,且满足:a1+a2+a3=6,a5=5;数列{bn}满足:bn-bn-1=an-1(n≥2,n∈N﹡),b1=1.
(Ⅰ)求an和bn;
(Ⅱ)记数列cn=
(n∈N*),若{cn}的前n项和为Tn,求Tn. 1
bn+2n
答
(Ⅰ)∵a1+a2+a3=6,a5=5;,∴
可得a1=1,d=1,…(2分)
3a1+3d=6
a1+4d=5
∴an=n (3分)
又bn-bn-1=an-1=n-1,(n≥2,n∈N*),b1=1,
∴当n≥2时,
bn=(bn-bn-1)+(bn-1-bn-2)+…+(b2-b1)+b1
=(n-1)+(n-2)+(n-3)+…+(2-1)+1
=
+1n(n−1) 2
=
,…(4分)
n2−n+2 2
又b1=1适合上式,…(5分)
∴bn=
. …(6分)
n2−n+2 2
(Ⅱ)∵cn=
=1
bn+2n
=2
n2+3n+2
=2(2 (n+1)(n+2)
−1 n+1
),…(8分)1 n+2
∴Tn=2(
−1 2
)+2(1 3
−1 3
)+2(1 4
−1 4
)+…+2(1 5
−1 n+1
)1 n+2
=2(
−1 2
)=1-1 n+2
=2 n+2
.…(12分)n n+2