数列1,(1+2),(1+2+22),…,(1+2+22+…+2n-1)…的前n项和为( )A. 2n-1B. n•2n-nC. 2n+1-nD. 2n+1-2-n
问题描述:
数列1,(1+2),(1+2+22),…,(1+2+22+…+2n-1)…的前n项和为( )
A. 2n-1
B. n•2n-n
C. 2n+1-n
D. 2n+1-2-n
答
∵1+2+22+…+2n-1=
=2n-11×(1−2n) 1−2
∴数列的前n项和为:1+(1+2)+(1+2+22)+…+(1+2+22+…+2n-1)
=(21-1)+(22-1)+(23-1)+…+(2n-1)
=21+22+23+…+2n-n
=
−n=2n+1-2-n2(1−2n) 1−2
故选D
答案解析:由1+2+22+…+2n-1=
=2n-1可知,数列的前n项和为:(21-1)+(22-1)+(23-1)+…+(2n-1)=21+22+23+…+2n-n=1×(1−2n) 1−2
−n=2n+1-2-n2(1−2n) 1−2
考试点:数列的求和.
知识点:本题为数列的求和问题,求出数列的通项公式并应用到数列中是解决问题的关键,属中档题.