已知数列{an}满足:a1=1;an+1-an=1,n∈N*.数列{bn}的前n项和为Sn,且Sn+bn=2,n∈N*.(1)求数列{an}、{bn}的通项公式;(2)令数列{cn}满足cn=an•b

题目简介

已知数列{an}满足:a1=1;an+1-an=1,n∈N*.数列{bn}的前n项和为Sn,且Sn+bn=2,n∈N*.(1)求数列{an}、{bn}的通项公式;(2)令数列{cn}满足cn=an•b

题目详情

已知数列{an}满足:a1=1;an+1-an=1,n∈N*.数列{bn}的前n项和为Sn,且Sn+bn=2,n∈N*
(1)求数列{an}、{bn}的通项公式;
(2)令数列{cn}满足cn=an•bn,求其前n项和为Tn
题型:解答题难度:中档来源:宁国市模拟

答案

(1)由已知a1=1;an+1-an=1,n∈N*
∴数列{an}为等差数列,首项为1,公差为1.
∴其通项公式为an=n…(3分)
∵Sn+bn=2,∴Sn+1+bn+1=2,
两式相减,化简可得
bn+1
bn
=class="stub"1
2

∴数列{bn}为等比数列,
又S1+b1=2,
∴b1=1,
bn=class="stub"1
2n-1
…(7分)
(2)由已知得:cn=n•class="stub"1
2n-1

Tn=1+class="stub"2
2
+class="stub"3
22
+…+class="stub"n
2n-1

class="stub"1
2
Tn=class="stub"1
2
+class="stub"2
22
+class="stub"3
23
+…+class="stub"n-1
2n-1
+class="stub"n
2n

class="stub"1
2
Tn=1+class="stub"1
2
+class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n-1
-class="stub"n
2n
=
1-class="stub"1
2n
1-class="stub"1
2
-class="stub"n
2n
=2(1-class="stub"1
2n
)-class="stub"n
2n
…(11分)
Tn=4(1-class="stub"1
2n
)-class="stub"n
2n-1
=4-class="stub"2+n
2n-1
…(13分)

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