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=

1
bn+2n
(n∈N*),若{cn}的前n项和为Tn,求Tn

(Ⅰ)∵a1+a2+a3=6,a5=5;,∴

3a1+3d=6
a1+4d=5
可得a1=1,d=1,…(2分)
∴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
=
n(n−1)
2
+1

=
n2−n+2
2
,…(4分)
又b1=1适合上式,…(5分)
∴bn=
n2−n+2
2
. …(6分)
(Ⅱ)∵cn=
1
bn+2n
=
2
n2+3n+2
=
2
(n+1)(n+2)
=2(
1
n+1
1
n+2
)
,…(8分)
∴Tn=2(
1
2
1
3
)+2(
1
3
1
4
)+2(
1
4
1
5
)+…+2(
1
n+1
1
n+2
)

=2(
1
2
1
n+2
)
=1-
2
n+2
=
n
n+2
.…(12分)