已知数列{an}满足a1=2,an+1=2(1+1/n)^2an,求an的通项公式

a[n+1]=2(1+1/n)^2*a[n]
a[n+1]=2*(n+1)^2/n^2*a[n]
a[n+1]/(n+1)^2=2*a[n]/n^2
即 a[n+1]/(n+1)^2 是以 an/1=2为首项,2为公比的等比数列,所以
a[n+1]/(n+1)^2= 2* 2^n=2^(n+1)
即 a[n]=n^2*2^n
验证n=1时 a1=1^2*2^1=2也成立
所以 a[n]=n^2*2^n