Show that 3- 3 1- 3- 4 2- 3- 5 3- - 3 n-2 n- - - 6 2 e 3 -1 5 e 3 -2-

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Question

Show that
$\frac{3 \cdot 3}{1 +} \frac{3 \cdot 4}{2 +} \frac{3 \cdot 5}{3 +} \dots \frac{3 (n + 2)}{n +} \dots = \frac{6 (2 e^{3} + 1)}{5 e^{3} - 2}$ .

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Solution

$u_{n} = n u_{n - 1} + 3 (n + 2) u_{n - 2};$
$u_{n} - (n + 3) u_{n - 1} = - 3 {u_{n - 1} - (n + 2) u_{n - 2}};$
...........................................
$u_{3} - 6 u_{2} = - 3 (u_{2} - 5 u_{2})$
By multiplication;
$u_{n} - (n + 3) u_{n - 1} = {(- 3)}^{n - 2} (u_{2} - 5 u_{1})$
Now,
$p_{1} = 9, q_{1} = 1, p_{2} = 18, q_{2} = 14,$
$p_{n} - (n + 3) p_{n - 1} = {(- 1)}^{n - 1} 3^{n + 1}$
$\frac{p_{n}}{(n + 3)!} - \frac{p_{n - 1}}{(n + 2)!} = \frac{{(- 1)}^{n - 1} 3^{n + 1}}{(n + 3)!}$
.....................
$\frac{p_{2}}{5!} - \frac{p_{1}}{4!} = \frac{(- 1) 3^{2}}{5!}$
By addition;
$\frac{p_{n}}{(n + 3)!} = \frac{3^{2}}{4!} - \frac{3^{3}}{5!} + \frac{3^{4}}{6!} - . . . . .$
$= \frac{1}{9} (\frac{3^{4}}{4!} - \frac{3^{5}}{5!} + \frac{3^{6}}{6!} - . . . . .);$
and $\frac{q_{n}}{(n + 3)!} - \frac{1}{4!} = \frac{3^{2}}{5!} - \frac{3^{3}}{6!} + \frac{3^{4}}{7!} - . . . . .$
$= \frac{1}{27} (\frac{3^{5}}{5!} - \frac{3^{6}}{6!} + \frac{3^{7}}{7!} - . . . . .)$
$e^{- 3} = 1 - 3 + \frac{3^{2}}{2!} - \frac{3^{3}}{3!} + . . . . . .$
$= - 2 + \frac{3^{4}}{4!} - \frac{3^{5}}{5!} + \frac{3^{6}}{6!} - . . . . . .$
$∴ \frac{p_{n}}{q_{n}} = \frac{1}{9} (e^{- 3} + 2) \div \frac{1}{27} (\frac{27}{4!} - 2 + \frac{3^{4}}{4!} - e^{- 3})$
$\frac{(e^{- 3} + 2)}{- 2 e^{- 3}} = \frac{6 (2 e^{3} + 1)}{5 e^{3} - 2}$

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Show that3⋅31+3⋅42+3⋅53+⋯3(n+2)n+⋯=6(2e3+1)5e3−2.

Show that
$\frac{3 \cdot 3}{1 +} \frac{3 \cdot 4}{2 +} \frac{3 \cdot 5}{3 +} \dots \frac{3 (n + 2)}{n +} \dots = \frac{6 (2 e^{3} + 1)}{5 e^{3} - 2}$ .