1.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
2.d=4
a1=S1
所以a1=1/8(a1+2)^2
(a1-2)^2=0
a1=2
an=2+4(n-1)=4n-2
bn=(t-1)^2[(4(n+2)-2]=(t-1)^2(4n+6)
设kn=4n+6(是等差数例,k1=10,d=4)
Tn=(t-1)^2[kn等差数例的和]接下去自己算