Question

Show that
$\frac{2}{1-}\frac{3}{5-}\frac{8}{7-}\cdots \frac{n^2-1}{2n+1}=\frac{n(n+3)}{2}$.

Answer 1

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

$u_n-n{u}_{n-1}=(n+1)\{u_{n-1}-(n-1)u_{n-2}\}$

...............................................

$u_3-3u_2=4(u_2-u_1)$

By multiplication we get

$u_n-n{u}_{n-1}=(n+1)n.....4(u_2-2u_1)$

Now, $p_1=2,q_1=1;p_2=10,q_2=2;$

$\therefore p_n-n{p}_{n-1}=(n+1)!$

$q_n-n{q}_{n-1}=0$

$q_n=n{q}_{n-1}=n(n-1)q_{n-2}=n!$

$\frac{p_n}{n!}-\frac{p_{n-1}}{(n-1)!}=n+1;.....\frac{p_1}{2!}-\frac{p_1}{1!}=3;$

By addition

$\frac{p_n}{n!}-2=3+4+.....+(n+1)$

$\frac{p_n}{n!}=\frac{n(n+3)}{2}$

$\frac{p_n}{q_n}=\frac{n(n+3)}{2}$