Solving a recurrence relation with n squared

Question

I have trouble solving the following recurrence:

$$a_{1}=1, a_{n}=a_{n-1}\cdot n^{2}$$ for $n>1$. It seems somewhat untypical to me, could you give me some general advice on dealing with such examples?

Answer 1

$a_2=2^2$, $a_3= 2^23^2$. In general, $a_n = (n!)^2$.

You can check by induction - true for $n=1$. If true for $n$, $a_{n+1}= (n!)^2(n+1)^2 = ((n+1)!)^2$.

Answer 2

$$\begin{align} a_n&=n^2 a_{n-1}\\ &=n^2(n-1)^2a_{n-2}\\ &=n^2(n-1)^2(n-2)^2a_{n-3}\\ &\vdots\\ &=n^2(n-1)^2(n-2)^2\cdots 2^2\overbrace{a_1}^{=1}\\ &=[n(n-1)(n-2)\cdots 2\cdot 1]^2\\ &=(n!)^2\qquad\blacksquare\end{align}$$

Answer 3

Alteratively to the excellent answer by @Vaneet, you can take the logarithm and define $\ln(a_n)=b_n$ to turn the product in a sum.

$$\ln(a_n)=\ln(a_{n-1})+2\ln(n),$$ or

$$b_n=b_{n-1}+2\ln(n).$$

With $b_0=0$, the solution of this classical recurrence is

$$b_n=2\sum_{k=1}^n\ln(k)$$ i.e.

$$a_n=\left(\prod_{k=1}^nk\right)^2.$$