因为:a1,a3,a6成等比数列
所以:a3^2 =a1 × a6
(a1+2d)^2=a1×(a1+5d)
a1^2 + 4a1 × d + 4 × d^2 = a1^2 + 5 × a1 × d
得到:
a1×d = 4×d×d
又,d不为0
所以:
d = a1/4 = 1/2
所以
Sn = a1+(n-1)d/2=(n^2 + 7n)/4
因为:a1,a3,a6成等比数列
所以:a3^2 =a1 × a6
(a1+2d)^2=a1×(a1+5d)
a1^2 + 4a1 × d + 4 × d^2 = a1^2 + 5 × a1 × d
得到:
a1×d = 4×d×d
又,d不为0
所以:
d = a1/4 = 1/2
所以
Sn = a1+(n-1)d/2=(n^2 + 7n)/4