(3a的n+2次方+a的n+1次方)÷(-三分之一的n-1次方)
问题描述:
(3a的n+2次方+a的n+1次方)÷(-三分之一的n-1次方)
答
[3a^(n+2)+a^(n+1)]÷[-1/3^(n-1)]
=[3a^(n+2)+a^(n+1)]÷[-3^(1-n)]
=3a^[(n+2)-(1-n)]+a^[(n+1)-(1-n)]
=3a^(2n+1)+a^2n