若数列{an}满足a1=12,a1+a2+…+an=n2an,则数列{an}的前60项和为______.-数学

题目简介

若数列{an}满足a1=12,a1+a2+…+an=n2an,则数列{an}的前60项和为______.-数学

题目详情

若数列{an}满足a1=
1
2
,a1+a2+…+an=n2an,则数列{an}的前60项和为______.
题型:填空题难度:中档来源:不详

答案

∵数列{an}的前n项的和Sn=a1+a2+…+an,∴Sn=n2an,
当n≥2时,Sn-1=(n-1)2an-1,两式相减得an=n2an-(n-1)2an-1,
即(n2-1)an=(n-1)2an-1,故
an
an-1
=class="stub"n-1
n+1

an
a1
=
a2
a1
×
a3
a2
×
a4
a3
×…×
an
an-1
=class="stub"1
3
×class="stub"2
4
×…×class="stub"n-2
n
×class="stub"n-1
n+1
=class="stub"2
n(n+1)

结合a1=class="stub"1
2
,可得an=class="stub"1
n(n+1)

当n=1时,也满足上式,故an=class="stub"1
n(n+1)
对任意n∈N+成立,
可得an=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

因此,数列数列{an}的前n项和为Sn=(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
4
)+…+(class="stub"1
n
-class="stub"1
n+1
)=1-class="stub"1
n+1
=class="stub"n
n+1

∴{an}的前60项和为class="stub"60
61

故答案为:class="stub"60
61

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