Question

关于数列的通项公式

求笫一问

Answer 1

an = a1+(n-1)d
Sn = a1+a2+...+an
= (2a1+(n-1)d)n/2

√Sn = √a1 + (n-1)d
n=2
√(2a1+ d) = √a1 + d
a1+ d = 2√a1d + d^2 (1)
n=3
√(3a1+ 3d) = √a1 + 2d
2a1+3d = 4√a1d + 4d^2 (2)
(2)-2(1)
d=2d^2
d(2d-1)=0
d=1/2

from (1)
a1+ 1/2 = √a1 +1/4
a1+ 1/4 =√a1
a1^2+ (1/2)a1+1/16 = a1
16a1^2- 8a1+1 =0
(4a1-1)^2=0
a1=1/4

an = 1/4 + (n-1)(1/2)
= (2n-1)/4