Question

已知等差数列{an}的首项a1=1,且对于n∈N*,S2n/Sn为常数,求数列{an}的通项公式

Answer 1

设公差 = d

an = a1 + (n-1) d = 1+ (n-1)d
Sn = (1 + 1+(n-1)d) * n /2 = (2+(n-1)d) n/2

a2n = a1 + (2n-1)d = 1 +(2n-1)d
S2n = (1 + 1+(2n-1)d) * (2n)/2 =(2+(2n-1)d) *(2n) /2

S2n/ Sn = (2+(2n-1)d)*2 / (2+(n-1)d)
S2/S1 = (2+(2-1)d) / a1 = 2+d
所以 n>1 时
(2+(2n-1)d)*2 / (2+(n-1)d) = 2+d

(2+(2n-1)d)*2 = (2+d) (2+(n-1)d)

整理得
(n-1)*d^2 -2nd +2d =0
(n-1)d(d-2) =0
所以
d = 0 或 d=2

{an}的通项公式为

an = a1 + (n-1) *0 = 1
或

an = a1 + (n-1) *2 = 1 + 2n -2 = 2n -1