已知数列{an}中,a1=1,a2n+1+an2+1=2(an+1an+an+1-an),求数列1a1a2,1a2a3,…,1anan+1,…的前n项和Sn.-数学

题目简介

已知数列{an}中,a1=1,a2n+1+an2+1=2(an+1an+an+1-an),求数列1a1a2,1a2a3,…,1anan+1,…的前n项和Sn.-数学

题目详情

已知数列{an}中,a1=1,a2n+1+an2+1=2(an+1an+an+1-an),求数列
1
a1a2
1
a2a3
,…,
1
anan+1
,…
的前n项和Sn
题型:解答题难度:中档来源:不详

答案

∵an+12+an2+1=2(an+1an+an+1-an)
∴an+12-2an+1•an+an2-2(an+1-an)+1=0
∴(an+1-an)2-2(an+1-an)+1=0
∴(an+1-an-1)2=0
∴an+1-an=1∴{an}为等差数列
∴an=a1+(n-1)•1=n
∴Sn=class="stub"1
a1a2
+class="stub"1
a2a3
+…+class="stub"1
anan+1

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

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

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