函数f(x)=x3,在等差数列{an}中,a3=7,a1+a2+a3=12,记Sn=f(3an+1),令bn=anSn,数列{bn}的前n项和为Tn(1)求{an}的通项公式和Sn(2)求证Tn<13

题目简介

函数f(x)=x3,在等差数列{an}中,a3=7,a1+a2+a3=12,记Sn=f(3an+1),令bn=anSn,数列{bn}的前n项和为Tn(1)求{an}的通项公式和Sn(2)求证Tn<13

题目详情

函数f(x)=x3,在等差数列{an}中,a3=7,a1+a2+a3=12,记Sn=f(
3an+1
)
,令bn=anSn,数列{bn}的前n项和为Tn
(1)求{an}的通项公式和Sn
(2)求证Tn
1
3
题型:解答题难度:中档来源:不详

答案

(1)设数列{an}的公差为d,
∵a3=7,a1+a2+a3=12,
∴a1+2d=7,3a1+3d=12
解得a1=1,d=3,∴an=3n-2
∵f(x)=x3
Sn=f(
3an+1
)
=an+1=3n+1             (6分)
(2)证明:∵bn=anSn=(3n-2)(3n+1)
class="stub"1
bn
=class="stub"1
(3n-2)(3n+1)
=class="stub"1
3
(class="stub"1
3n-2
-class="stub"1
3n+1
)
Tn=class="stub"1
b1
+class="stub"1
b2
+…+class="stub"1
bn
=class="stub"1
3
(1-class="stub"1
4
+class="stub"1
4
-class="stub"1
7
+…+class="stub"1
3n-2
-class="stub"1
3n+1
)

Tn=class="stub"1
3
(1-class="stub"1
3n+1
)<class="stub"1
3
(12分)

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