已知等差数列an的前n项和为sn 且满足Sn=n²+n,则通项公式an=?
问题描述:
已知等差数列an的前n项和为sn 且满足Sn=n²+n,则通项公式an=?
答
因为 S(n-1) = (n-1)^2 + (n-1),所以:
an = Sn - S(n-1) = (n^2 + n) - [(n-1)^2 + (n-1)]
= n^2 - (n-1)^2 + 1
= 2n