题目内容

设等差数列{an}的前n项和为Sn,且a3=6,S10=110.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设数列{bn}前n项和为Tn,且Tn=1-(
2
2
)an
,令cn=anbn(n∈N*).求数列{cn}的前n项和Rn
(Ⅰ)设等差数列{an}的公差为d,
∵a3=6,S10=110.
∴a1+2d=6,10a1+
10×9
2
d=110

解得a1=2,d=2,
∴数列{an}的通项公式an=2+(n-1)•2=2n;
(Ⅱ)∵Tn=1-(
2
2
)an=1-(
2
2
)2n=1-(
1
2
)n

当n=1时,b1=T1=1-(
2
2
)2
=
1
2

当n≥2时,bn=Tn-Tn-1=1-(
1
2
)n-[1-(
1
2
)n-1]
=(
1
2
)n

且n=1时满足,
∴数列{an}的通项公式为bn=(
1
2
)n

又an=2n,
cn=
2n
2n
=
n
2n-1

Rn=
1
20
+
2
21
+
3
22
+…+
n
2n-1

1
2
Rn=
1
2
+
2
22
+
3
23
+…+
n
2n

两式相减得:
1
2
Rn=
1
20
+
1
21
+
1
22
+…+
1
2n-1
-
n
2n
=
1-
1
2n
1-
1
2
-
n
2n
=2-
n+2
2n

Rn=4-
n+2
2n-1
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