f(x)=ax²+bx+c過(0,1)和(1,4)代入
c=1
a+b+c=4
∴b=3-a
則f(x)=ax²+（3-a）x+1,與Y=4x相切,則聯(lián)立二方程得
即ax²-（a+1）x+1=0
有△=[-（a+1）]²-4a=0即（a-1）=0
∴a=1
則b=2
∴f(x)=x²+2x+1
(2)
(2)
a1=f(1)=4
當(dāng)n≥2時an=f(n)-f(n-1)=(n^2+2n+1)-[(n-1)^2+2(n-1)+1]=2n+1
所以an=4 (n=1)
an= 2n+1 (n≥2)
令bn=1/an*a(n+1)吧,如果是,則有:bn=1/(2n+1)*(2n+3)=1/2[1/(2n+1)-1/(2n+3)]
Sn=1/a1*a2+1/a2*a3+...+1/(an*a(n+1))
=1/4*5+1/2[1/5-1/7+1/7-1/9+.+1/(2n+1)-1/(2n+3)]
=1/20+1/2[1/5-1/(2n+3)]
=1/20+1/10-1/2(2n+3)
=3/20-1/(4n+6)