已知:數(shù)列{an}前n項的和Sn=n^2+[(3/2)n]+2.求通項公式.

熱心網(wǎng)友

a(n)=s(n)-s(n-1)=n^2+[(3/2)n]+2-(n-1)^2-[(3/2)(n-1)]-2 =2n+1/2