求和:Sn=1+(1+1/2)+(1+1/2+1/4)+[1+1/2+1/4+…+(1/2)n-1].

问题描述:

求和:Sn=1+(1+

1
2
)+(1+
1
2
+
1
4
)+[1+
1
2
+
1
4
+…+(
1
2
n-1].

∵1+

1
2
+
1
4
+…+(
1
2
n-1=
1−(
1
2
)n
1−
1
2
=2−
1
2n−1

Sn=2n−(1+
1
2
+
1
22
+…+
1
2n−1
)
=2n-
1−
1
2n
1−
1
2
=2n-2+
1
2n−1