Question

已知{an}是公差不为零的等差数列a1=1且a1,a3,a9成等比数列．求数列{an}的通项；求数列{2an}的前n项Sn和

Answer 1

设等差为d
a1,a3,a9成等比数列
1*（1+8d）=（1+2d)^2
d=1
数列{an}的通项an=n
{2an}的前n项Sn=2*(1+n)*n/2=n^2+n

Answer 2

a1,a3,a9成等比数列 (A3)^2=A1*A9=A9 A3=A1+2d=1+2d A9=A=2*(A1+An)*n/2 =n(n+1)

Answer 3

(a1+2d)^2=a1+8d
a1^2+4da1+4d^2-a1-8d=0
1+4d+4d^2-1-8d=0
4d^2-8d=0
d=2
an=1+(n-1)*2=1+2n-2=2n-1

Answer 4

A3*A3=A1*A9 得到D=1 得AN=N 2AN=2+2(N-1) SN=(A1+AN)*N/2 SN=N(N+1) OK