Question

Show that
$\frac{2}{2-}\frac{3}{3-}\frac{4}{4-}\cdots \frac{n+1}{n+1-}\frac{n+2}{n+2}=1+1+2!+3!+\cdots +n!$.

Answer 1

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

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

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

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

$\frac{u_3}{4}-\frac{u_2}{3}=2(\frac{u_2}{3}-\frac{u_1}{2})$

By multiplication we get

$\frac{u_{n+1}}{n+2}-\frac{u_n}{n+1}=n!(\frac{u_2}{3}-\frac{u_1}{2})$

$p_1=2,q_1=2,p_2=6,q_2=3,$

$\therefore \frac{q_{n+1}}{n+2}-\frac{q_n}{n+1}=0$

$\frac{q_{n+1}}{n+2}=\frac{q_n}{n+1}$

$=\frac{q_{n-1}}{n}=\frac{q_1}{2}=1;$

$\frac{p_{n+1}}{n+2}-\frac{p_n}{n+1}=n!....\frac{p_2}{3}-\frac{p_1}{2}=1!$

$\frac{p_{n+1}}{n+2}=1+1!+2!+3!+.....+n!;$

$\frac{p_{n+1}}{q_{n+1}}=1+1!+2!+3!+.....+n!$