数列{an}满足an+1-an=1/2(n∈N*),a1=1/2,Sn是数列{an}的前n项和,则S100=_.

问题描述:

数列{an}满足an+1-an=

1
2
(n∈N*),a1=
1
2
,Sn是数列{an}的前n项和,则S100=______.

由an+1-an=

1
2
(n∈N*)知:数列{an}为以
1
2
为公差的等差数列,
又a1=
1
2

所以S100=100×
1
2
+
100×99
2
×
1
2
=2525,
故答案为:2525.