设等差数列:2,a+2,3a,…的前n项和为Sn,则1S1+1S2+…+1S100的值是______.-数学

题目简介

设等差数列:2,a+2,3a,…的前n项和为Sn,则1S1+1S2+…+1S100的值是______.-数学

题目详情

设等差数列:2,a+2,3a,…的前n项和为Sn,则
1
S1
+
1
S2
+…+
1
S100
的值是______.
题型:填空题难度:中档来源:不详

答案

∵等差数列前三项为2,a+2,3a,
∴2×(a+2)=2+3a,
∴a=2,
公差d=4-2=2
所以等差数列2,4,6,…的前n项和Sn=
n(2+2n)
2
,即Sn=n(n+1)
于是 class="stub"1
Sn
=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

class="stub"1
S1
+class="stub"1
S2
+…+class="stub"1
S100
=(1-class="stub"1
2
)+( class="stub"1
2
-class="stub"1
3
)+( class="stub"1
3
-class="stub"1
4
)+…+( class="stub"1
100
-class="stub"1
101
)=1-class="stub"1
101
=class="stub"100
101

故答案为:class="stub"100
101

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