设等差数列:2,a+2,3a,…的前n项和为Sn,则
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∵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=
,即Sn=n(n+1)n(2+2n) 2
于是
=1 Sn
=1 n(n+1)
-1 n
,1 n+1
则
+1 S1
+…+1 S2
=(1-1 S100
)+( 1 2
-1 2
)+( 1 3
-1 3
)+…+( 1 4
-1 100
)=1-1 101
=1 101 100 101
故答案为:
.100 101