等比数列求通项~

题目

等比数列求通项~
数列{an}中,a1=1,a(n+1)=3an+n求通项

答案

a(n+1)=3an+n=3an+3n/2-n/2=3an+3n/2 - (n+1)/2 + 1/2
a(n+1)+(n+1)/2 = 3[a(n)+n/2] + 1/2 = 3[a(n)+n/2] + 3/4 - 1/4
a(n+1)+(n+1)/2 + 1/4 = 3[a(n)+n/2+1/4]
所以{a(n)+n/2+1/4}是首项为a(1)+1/2+1/4=1+3/4=7/4,公比为3的等比数列
于是
a(n)+n/2+1/4 = (7/4)*3^(n-1)
a(n) = (7/4)*3^(n-1)-n/2-1/4