已知正项数列{an},其前项和满足10Sn=an2（2是平方）+5an+6,求数列{an}的通项an

解析：∵10Sn=（an）^2+5an+6
∴10Sn-1=（an-1）^2+5an-1+6
则,10(Sn-Sn-1)=10an
=（an）^2+5an-（an-1）^2-5an-1
∴（an）^2-（an-1）^2=5an+5an-1
即（an+an-1)(an-an-1)=5(an+an-1)
∵an+an-1＞0
∴an-an-1=5,
又10a1=(a1)^2+5a1+6,
解得a1=2,或a1=3
∴{an}是首相a1=2,或a1=3,公差d=5的等差数列.
∴an=2+5（n-1)=5n-3
或an=3+5(n-1)=5n-2

10s1=(a1)^2+5a1+6
10a1=(a1)^2+5a1+6
(a1)^2-5a1+6=0
(a1-2)(a1-3)=0
a1=2 或a1=3
10Sn=(an)^2）+5an+6
10S(n-1)=[a(n-1)]^2+5a(n-1)+6
10an=10Sn-10S(n-1)
=(an)^2）+5an-[a(n-1)]^2...

10Sn = an^2 + 5an + 6 ①
10S(n-1) = a(n-1)^2 + 5a(n-1) + 6 ②
①-②后整理，
[an+a(n-1)]*[an-a(n1)-5]=0
∵正项数列。
∴必有an-a(n-1)+5=0
∴等差数列，且d=5
取n=1，则a1^1-5a1+6=0得a1=2或3
剩下的就简单了- -

10Sn=an^2+5an+6
10S(n-1)=a(n-1)^2+5a(n-1)+6
10an=10Sn-10S(n-1)=an^2+5an-a(n-1)^2+5a(n-1)
[an+a(n-1)]·[an-a(n-1)-5]=0
an>0
所以an-a(n-1)＝5
为等差数列,d=5。
10Sn=an^2+5an+6;n=1时，10a1=a1^2+5a1+6
a1=2或3
an=2+5(n-1)=5n-3
或an=5n-2