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}是等差数列.
问题描述：
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}是等差数列.
答
a = S - S
= (1/8) * (a + 2)^2 - (1/8)*(a +2)^2
= (1/8)*a^2 + (1/2)*a - (1/8)*a^2 - (1/2)*a
移项
0 = (1/8)*a^2 - (1/2)*a - (1/8)*a^2 - (1/2)*a
分解因式
(1/8)*[a^2 - a^2] - (1/2)*(a + a) = 0
(1/4)*(a + a)(a-a) - (a+a) = 0
(a + a)*[(1/4)(a - a) - 1] = 0
因为 a 大于0,所以 a + a 不为0
所以
(1/4)*(a - a) - 1 = 0
a - a = 4
因此 an 是公差为4的等差数列