已知数列an的通项公式为an=n+12,设Tn=1a1•a3+1a2•a4+…+1an•an+2,求Tn.-数学

题目简介

已知数列an的通项公式为an=n+12,设Tn=1a1•a3+1a2•a4+…+1an•an+2,求Tn.-数学

题目详情

已知数列an的通项公式为an=
n+1
2
,设Tn=
1
a1a3
+
1
a2a4
+…+
1
anan+2
,求Tn
题型:解答题难度:中档来源:不详

答案

an=class="stub"n+1
2

class="stub"1
anan+2
=class="stub"4
(n+1)(n+3)
=2(class="stub"1
n+1
-class="stub"1
n+3
)

Tn=2(class="stub"1
2
 -class="stub"1
4
+class="stub"1
3
-class="stub"1
5
+…+class="stub"1
n+1
-class="stub"1
n+3
)

=2(class="stub"1
2
+class="stub"1
3
-class="stub"1
n+2
- class="stub"1
n+3
)=
5n2+25n+24
3(n+2)(n+3)

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