Question

已知数{an}的前n项和公式为Sn=n^2+3n，求证：{an}是等差数列

步骤

Answer 1

Sn-Sn-1=an=n^2+3n-[(n-1)^2+3(n-1)]=n^2+3n-(n^2-2n+1+3n-3)=n^2+3n-(n^2+n-2)=n^2+3n-n^2-n+2=2n+2
=4+(n-1)X2
所以an为首项为4
公差为2的等差数列

Answer 2

n=1时,a1=S1=1�0�5+3×1=4n>1时Sn=n�0�5+3n①S(n-1)=(n-1)�0�5+3(n-1)=n�0�5+n-2②①-②得 Sn-S(n-1)=(n�0�5+3n)-(n�0�5+n-2)=2n+2∴an=Sn-S(n-1)=2n+1,n=1时也符合∴{an}通项为an=2n+2a(n+1)-an=2(n+1)+2-(2n+2)=2,即从第二项起,每一项与前一项差为常数∴{an}是等差数列

Answer 3

a1=s1=4n>=2时,an=Sn-Sn-1=n^2+3n-(n-1)^2-3(n-1)=2n+2所以，an=2n+2为等差