Question

已知等差数列〔an〕的前四项和为10,且a2a3a7成等比数列。

〔1〕求通项公式an(2)设bn=2的an次方，求数列bn的前n项和sn

Answer 1

设首项为a1,公差为d
a1+(a1+d)+(a1+2d)+(a1+3d)
=4a1+6d=10
a1=(5-3d)/2

a3=a1+2d,a7=a1+6d
由a2a3a7成等比数列

(a1+2d)^2=(a1+d)(a1+6d)

a1^2+4a1d+4d^2=a1^2+7a1d+6d^2
3a1d+2d^2=0
把a1=(5-3d)/2代入解得
a1=-2,d=3
所以an=-2+3(n-1) n=1,2,3,4……

b1=2^-2,b2=2^1,b3=2^4,b4=2^7
bn是一个公比为2^3的等比数列

sn=b1(1-q^n)/(1-q)
=2^(-2)(1-8^n)/(1-8)
=(8^n-1)/28