已知数列{an}为首项a1≠0,公差为d≠0的等差数列,求Sn=1a1a2+1a2a3+…+1anan+1.-数学

题目简介

已知数列{an}为首项a1≠0,公差为d≠0的等差数列,求Sn=1a1a2+1a2a3+…+1anan+1.-数学

题目详情

已知数列{an}为首项a1≠0,公差为d≠0的等差数列,求Sn=
1
a1a2
+
1
a2a3
+…+
1
anan+1
题型:解答题难度:中档来源:不详

答案

由等差数列的性质可得,class="stub"1
anan+1
=class="stub"1
d
(class="stub"1
an
-class="stub"1
an+1
)

∴Sn=class="stub"1
d
[(class="stub"1
a1
-class="stub"1
a2
)+(class="stub"1
a2
-class="stub"1
a3
)+…+(class="stub"1
an
-class="stub"1
an+1
)]=class="stub"1
d
(class="stub"1
a1
-class="stub"1
an+1
)=class="stub"1
d
an+1-a1
a1an+1

=class="stub"n
a1(a1+nd)

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