Question

Show that
$\frac{1}{1-}\frac{4}{5-}\frac{9}{13-}\frac{64}{25-}\cdots \frac{{(n^2-1)}^2}{n^2+{(n+1)}^2}=\frac{(n+1)(n+2)(2n+3)}{6}$.

Answer 1

Here, $\frac{{(n^2-1)}^2}{n^2+{(n+1)}^2}$is the $(n+1)th$ component

$\therefore u_{n+1}=\{n^2+{(n+1)}^2\}{u}_n-{(n^2-1)}^2{u}_{n-1}$

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

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

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

Hence by multiplication we obtain

$u_{n+1}-n^2{u}_n=3^2.4^2.....{(n+1)}^2(u_2-u_1)$

Now, $p_1=1,q_1=1;p_2=5,q_2=1;$

$q_{n+1}-n^2{q}_n=0$

$q_{n+1}=n^2{q}_n$

$q_{n+1}=n^2{(n-1)}^2....1^2={(n!)}^2$

$p_{n+1}-n^2{p}_n=3^2.4^2.....{(n+1)}^{2}4$

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

$\frac{p_2}{{(1!)}^2}-\frac{p_2}{1}=2^2;$

By addition;-

$\therefore \frac{p_{n+1}}{{(n!)}^2}-1=2^2+3^2+.....+{(n+1)}^2$

$\frac{p_{n+1}}{{(n!)}^2}=1^2+2^2+.....+{(n+1)}^2$

$\frac{q_{n+1}}{{(n!)}^2}=1$

$\frac{p_{n+1}}{q_{n+1}}=\sum n^2=\frac{(n+1)(n+2)(2n+3)}{6}$