If a(n)=a(n-1)*(n-2)^2, what is the formula for a(n) without knowing a(n-1)?

Question

I know that the n-th term of a geometric sequence (with initial value 'a' and a common ratio 'r') can be calculated as: a_n=ar^(n-1). How do you calculate the nth term however, if 'r' itself is dependent on 'n' and includes a power? For example, if a_n=a_(n-1)*(n+2)², what is the formula for a_n without knowing a_(n-1)? What kind of sequence is this?

Answer 1

a_n=a_n-1(n+2)²

From here it follows that a_n/a_n-1=(n+2)² =r.

Geometric progression is a, ar, ar², ar³,..

 

Now, let a₀ be some constant.

a₁=a₀3^2.  

 

a₂=a₀(3*4)²

a₃=a₀(3*4*5)²

a₄=a₀(3*4*5*6)²

This is not a a geometric progression but  increasing (if a₀>0) and decreasing (if a₀<0) sequence, which doesn't converge

So, if you know a₀ then you can express a_n (without a_n-1)  as

a_n=a₀(3*4*...*(n+2))²=(a₀/4)*((n+2)!)²

But again this is not a geometric progression because in order to obtain next member (of the sequence) we multiply this member by a different number than it was for the previous one.