Find n!

Both the numerator and the denominator are arithmetic progressions:

• the numerator: $p_n = 4n + 1$

• the denominator: $q_n = 2n + 5$

$\text {By applying the well known formula for the sum } (S_n = \dfrac {a_1 + a_n }{2} × n ) \text { we get: }$

$\dfrac {S_ {p_n}}{S_{q_n}} = \dfrac { \frac {5 + 4n + 1 }{2} × n }{ \frac {7 + 2n + 5 }{2} × n } = \dfrac {4n + 6 }{2n + 12} = \dfrac {2n + 3 }{n + 6}$

Now, we can set up the following equation:

$\dfrac {2n + 3 }{n + 6} = \dfrac {5}{4}$

$8n + 12 = 5n + 30$

$3n = 18$

$n = \boxed {6}$