若数列｛an｝的前n项和Sn=n∧²+2n,求an.

a(1)=s(1)=1+2=3,
s(n) = n^2 + 2n,
s(n+1)=(n+1)^2 + 2(n+1),
a(n+1) = s(n+1)-s(n) = (n+1)^2 + 2(n+1) - n^2 - 2n = (2n+1) + 2 = 2n+3,
a(n) = 2(n-1) + 3 = 2n + 1.