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+1 2
+1 2
)+[1+1 4
+1 2
+…+(1 4
)n-1]. 1 2
答
∵1+
+1 2
+…+(1 4
)n-1=1 2
=2−1−(
)n
1 2 1−
1 2
,1 2n−1
∴Sn=2n−(1+
+1 2
+…+1 22
)=2n-1 2n−1
=2n-2+1−
1 2n 1−
1 2
.1 2n−1