a[n+1] = 3a[n] + 1.
也就是說,數(shù)列的每個(gè)數(shù)等于前面那個(gè)數(shù)的三倍加上1.
這個(gè)可以解出來通項(xiàng)
a[n+1] + 1/2 = 3a[n] + 3/2.
所以,a[n] + 1/2 是等比數(shù)列
a[n] = aq^n - 1/2
a[1]=0,1/2 = aq
a[2]=1,3/2 = aq^2
q = 3,a = 1/6
所以通項(xiàng) a[n] = (1/6) * 3^n - 1/2.