设数列{an}中,首相a1=3,点(√an+1,√an)(n=1,2,3,4.）均在直线上x-y-√3=0上

设数列{an}中,首相a₁=3,点(√an+1,√an)(n=1,2,3,4.）均在直线上x-y-√3=0上
（1）求数列{an}的通项公式
（2）求数列若bn=1/[(4an/3)-1],{bn }前n项的和Sn
(1)
∵ a₁= 3, √a₁= √3
∵√(an+₁) -√an = √3
√(an+₁) = √an + √3
∴ √a₂= √a₁+ √3 = 2√3
√a₃= √a₂+ √3 = 3√3
√a₄= √a₃+ √3 = 4√3
.
√an = √an-₁+ √3 = n√3

∴ an = 3n²（这就是通项公式）
(2)
bn = 1/(4an/3 - 1)
= 1/(4×3n²/3 - 1)
= 1/(4×n² - 1)
= 1/(2n - 1)(2n + 1)
= ½[1/(2n - 1) - 1/(2n + 1)]
Sn = b₁+ b₂+ b₃+ b₄+ . + bn
= ½[(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+.+(1/(2n-1)-1/(2n+1))]
= ½[(1-1/(2n+1)]
= ½[2n/(2n+1)]
= n/(2n+1)