数学设数列{an}中a1=1且an+1=an+3n+1求an

数学设数列{an}中a1=1且an+1=an+3n+1求an: 设数列{an}中a1=1且an+1=an+3n+1求an; 解：∵a(n+1)=a(n)+3n+1 = a(n) + (3/2)[n(n+1)-(n-1)n] + [(n+1)-n], a(n+1) - 3n(n+1)/2 - (n+1) = a(n) - 3(n-1)n/2 - n, ∴{a(n)-3(n-1)n/2 - n}是首项为a(1)-0-1=0的常数数列。有：a(n)-3(n-1)n/2 - n = 0, 得到：a(n) = 3(n-1)n/2 + n=n(3n-1)/2。