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

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

∵1+2+22+…+2n-1=

1×(1−2n)
1−2
=2n-1
∴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
=
2(1−2n)
1−2
−n
=2n+1-2-n
故选D