Solve recursion

Question

$a_n=n^2\times a_{(n-1)}$, for $ n>0, a_0=1$

Whats the correct formula to use in this case?

I solved problem with recursion before, but they had '$+$' instead of $'\times'$ so there is following formula to simply solve them: $ar^n+bnr^n$.

Is there something similar, but for multiplication instead?

Compute the first terms for $\sqrt{a_n}$. – Aphelli Jan 26 '19 at 23:29 — Aphelli, Jan 26 '19 at 23:29

score 2 · Answer 1 · answered Jan 26 '19 at 23:30

2

$a_n=(n!)^2$ using induction. Try to evaluate the first few terms on your own.

answered Jan 26 '19 at 23:30

abc...

4,904

This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n \cdot (n-1) \cdot \ldots \cdot 2 \cdot 1$... – Klaus Jan 26 '19 at 23:37

score 1 · Answer 2 · answered Jan 27 '19 at 00:37

Multiply these lines together: \begin{align} a_n&=n^2\times a_{n-1}\\ a_{n-1}&=(n-1)^2\times a_{n-2}\\ &\dots\\ a_2&=2^2\times a_{1}\\ a_1&=1^2\times a_{0}\\ \end{align} After canceling, we get $$ a_n = 1^2 \cdot 2^2 \cdots n^2 a_0 = (1 \cdot 2 \cdots n)^2 = (n!)^2 $$

Solve recursion

2 Answers2