分类：教育/科学-升学入学-高考
问题：已知数列{An}的前n项和Sn＝π（2n^2+n）/12求证：{? 
已知数列{An}的前n项和Sn＝π（2n^2+n）/12
求证：{An}是等差数列，并求公差与首项
回答：Sn=π(2n^2+n)/12=(π/12)*(2n^2+n)
所以：
a1=S1=(π/12)*(2+1)=π/4

S<n-1>=(π/12)*[2(n-1)^2+(n-1)]
=(π/12)*(2n^2-4n+2+n-1)
=(π/12)*(2n^2-3n+1)

则，S<n>-S<n-1>=(π/12)*(2n^2+n)-(π/12)*(2n^2-3n+1)
=(π/12)*(2n^2+n-2n^2+3n-1)
=(π/12)*(4n-1)
即，an=(π/12)*(4n-1)
所以，a<n-1>=(π/12)*[4(n-1)-1]=(π/12)*(4n-5)

所以，an-a<n-1>=(π/12)*(4n-1)-(π/12)*(4n-5)
=(π/12)*(4n-1-4n+5)
=π/3

则，数列an是以a1=π/4为首项，公差d=π/3的等差数列。