Question

已知数列{an}中，an>0,前n项和为Sn,且满足Sn=1/8(an+2)^2.求证数列{an}的通项公式

Answer 1

Sn=1/8(an+2)^2
S(n-1)=1/8[a(n-1)+2]^2
2式相减整理得到
[an + a(n-1)]*[(1/4)(an - an-1) - 1] = 0
又an＞0
所以1/4(an-an-1)-1=0
an-an-1=4
所以an是个公差为4的等差数列
a1=1/8(a1+2)^2
8a1=a1^2+4a1+4
(a1-2)^2=0
a1=2
所以an=2+(n-1)*4=4n-2

Answer 2

a<n> = S<n> - S<n-1>
= (1/8) * (a<n> + 2)^2 - (1/8)*(a<n-1> +2)^2
= (1/8)*a<n>^2 + (1/2)*a<n> - (1/8)*a<n-1>^2 - (1/2)*a<n-1>

移项
0 = (1/8)*a<n>^2 - (1/2)*a<n> - (1/8)*a<n-1>^2 - (1/2)*a<n-1>

分解因式
(1/8)*[a<n>^2 - a<n-1>^2] - (1/2)*(a<n> + a<n-1>) = 0
(1/4)*(a<n> + a<n-1>)(a<n>-a<n-1>) - (a<n>+a<n-1>) = 0

(a<n> + a<n-1>)*[(1/4)(a<n> - a<n-1>) - 1] = 0

因为 a<n> 大于0， 所以 a<n> + a<n-1> 不为0
所以
(1/4)*(a<n> - a<n-1>) - 1 = 0
a<n> - a<n-1> = 4

因此 an 是公差为4的等差数列