设Sn是数列{an}的前n项和,且点(n,Sn)在函数y=x2+2x上,(1)求数列{an}的通项公式;(2)已知bn=2n-1,Tn=1a1.b1+1a2.b2+…+1an.bn,求Tn.-数学

题目简介

设Sn是数列{an}的前n项和,且点(n,Sn)在函数y=x2+2x上,(1)求数列{an}的通项公式;(2)已知bn=2n-1,Tn=1a1.b1+1a2.b2+…+1an.bn,求Tn.-数学

题目详情

设Sn是数列{an}的前n项和,且点(n,Sn)在函数y=x2+2x上,
(1)求数列{an}的通项公式;
(2)已知bn=2n-1,Tn=
1
a1b1
+
1
a2b2
+…+
1
anbn
,求Tn
题型:解答题难度:中档来源:不详

答案

(1)由题意可得,Sn=n2+2n
当n=1时,a1=S1=3
当n≥2时,an=Sn-Sn-1=n2+2n-(n-1)2-2(n-1)=2n+1
而a1=3适合上式
∴an=2n+1
(2)∵bn=2n-1
class="stub"1
anbn
=class="stub"1
(2n-1)(2n+1)
=class="stub"1
2
(class="stub"1
2n-1
-class="stub"1
2n+1
)

∴Tn=class="stub"1
a1b1
+class="stub"1
a2b2
+…+class="stub"1
anbn

=class="stub"1
2
(1-class="stub"1
3
+class="stub"1
3
-class="stub"1
5
+…+class="stub"1
2n-1
-class="stub"1
2n+1
)

=class="stub"1
2
(1-class="stub"1
2n+1
)
=class="stub"n
2n+1

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