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$\bullet$ This is taken out of my notes and basically I have to fill in the blanks(1 and 2) but I'm stuck. Can't seem to come up with a general term by even by looking at these patterns. Would appreciate some help here!

Question:

A sequence $u_1,u_2,u_3...$ is given by $u_{r+1}=\frac{r+1}{r^2}u_r$ and $u_1=1$. Find an expression for $u_n$ in terms of n.

$u_2=\frac{2}{1^2}$

$u_3=\frac{3}{2^2}\cdot\frac{2}{1^2}$

$u_4=\frac{4}{3^2}\cdot\frac{3}{2^2}\cdot\frac{2}{1^2}$

  1. Therefore, $u_n=$ ....

Alternative method:

$u_n=\frac{(n-1)+1}{(n-1)^2}u_{n-1}$

$\;\;\;\;\;=\frac{n}{(n-1)^2}u_{n-1}$

$\;\;\;\;\;=\frac{n}{(n-1)^2}\frac{n-1}{(n-2)^2}u_{n-2}$

$\;\;\;\;\;=\frac{n}{(n-1)^2}\frac{n-1}{(n-2)^2}\frac{n-2}{(n-3)^2}u_{n-3}$

  1. Therefore $u_n=$ ....
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    $\begingroup$ Is it not $\frac{n}{(n-1)!}$ $\endgroup$ Oct 30 '18 at 14:29
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$$=\prod_{r=1}^m\dfrac{(n+1-r)}{(n-r)^2}u_{n-m}=u_{n-m}\dfrac n{(n-m)\prod_{r=1}^m(n-r)}$$

Set $m=n-1$

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