令bn = log an
则
b(n+1) = log a(n+1) = log 3an^4
= log 3 + 4log an
= log 3 + 4bn
可以解得
[b(n+1) + (log 3)/3 ] = 4[bn + (log 3)/3 ]
bn + (log 3)/3 = 4^(n-1)(b1 + (log 3)/3)
剩下的就是代入的工作了
令bn = log an
则
b(n+1) = log a(n+1) = log 3an^4
= log 3 + 4log an
= log 3 + 4bn
可以解得
[b(n+1) + (log 3)/3 ] = 4[bn + (log 3)/3 ]
bn + (log 3)/3 = 4^(n-1)(b1 + (log 3)/3)
剩下的就是代入的工作了