Question

设等差数列｛an｝的前n项和为Sn，且S4=4S2，a4=2a2+1，求数列｛an｝的通项公式及其

设等差数列｛an｝的前n项和为Sn，且S4=4S2，a4=2a2+1，求数列｛an｝的通项公式及其前n项和Sn

Answer 1

由等差数列前n项和公式：sn=na1+n(n-1)d/2,且
s4=4s2 可得 4a1+6d=4(2a1+d) 化简得 2a1=d (1)式
又由a4=2a2+1 可得 a1+3d=2(a1+d)+1 化简得 d-a1=1 (2)式
由（1）（2）两式可解得 a1=1,d=2
于是an=a1+(n-1)d=2n-1 sn=na1+n(n-1)d/2=n+n(n-1)*2=n^2

Answer 2

an = a1+(n-1)d

S4=4S2
(2a1+3d)2 = 4(2a1+d)
2a1=d (1)

a4=2a2+1
a1+3d = 2(a1+d)+1
7a1=6a1+1
a1=1
from (1) d=1/2
an = 1+ (1/2)(n-1) = (1/2)(n+1)

Answer 3

Answer 4

Answer 5