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-(
)an,令cn=anbn(n∈N*).求数列{cn}的前n项和Rn.
(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 |
(Ⅰ)设等差数列{an}的公差为d,
∵a3=6,S10=110.
∴a1+2d=6,10a1+
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-(
)an=1-(
)2n=1-(
)n,
当n=1时,b1=T1=1-(
)2=
,
当n≥2时,bn=Tn-Tn-1=1-(
)n-[1-(
)n-1]=(
)n,
且n=1时满足,
∴数列{an}的通项公式为bn=(
)n.
又an=2n,
∴cn=
=
,
∴Rn=
+
+
+…+
,
即
Rn=
+
+
+…+
,
两式相减得:
Rn=
+
+
+…+
-
=
-
=2-
,
∴Rn=4-
.
∵a3=6,S10=110.
∴a1+2d=6,10a1+
10×9 |
2 |
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 |
1 |
2 |
当n=1时,b1=T1=1-(
| ||
2 |
1 |
2 |
当n≥2时,bn=Tn-Tn-1=1-(
1 |
2 |
1 |
2 |
1 |
2 |
且n=1时满足,
∴数列{an}的通项公式为bn=(
1 |
2 |
又an=2n,
∴cn=
2n |
2n |
n |
2n-1 |
∴Rn=
1 |
20 |
2 |
21 |
3 |
22 |
n |
2n-1 |
即
1 |
2 |
1 |
2 |
2 |
22 |
3 |
23 |
n |
2n |
两式相减得:
1 |
2 |
1 |
20 |
1 |
21 |
1 |
22 |
1 |
2n-1 |
n |
2n |
1-
| ||
1-
|
n |
2n |
n+2 |
2n |
∴Rn=4-
n+2 |
2n-1 |
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