Question

数列an的前n项和Sn=n²+2n(N∈N+,n≥1),则数列通项an

Answer 1

Sn=n²+2n,Sn-1=(n-1)²+2(n-1)=n²-1.an=Sn-Sn-1=2n+1.验证a1=3，因此通项an=2n+1

Answer 2

Sn=n²+2n,Sn-1=(n-1)²+2(n-1),

Sn-Sn-1=an=2n+1,

当N=1时S1=3
an也=3

Answer 3

∵Sn=n²+2n

∴a(n+1)=s(n+1)-sn
=2n+3
=2(n+1)+1
∴an=2n+1

Answer 4

不有公式吗？？？
Sn=na1+dn(n-1)/2=d/2n²+（a1-d/2)n
题目中Sn=n²+2n(N∈N+,n≥1)
不就可以根据系数关系得出a1和d 的值
得，a1=3,d=2
∴an=2n+1