faind n terme 4,−2,((16)/9),−2,......


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Question Number 196053 by mathlove last updated on 17/Aug/23

$f a i n d n t e r m e$ $4, - 2, \frac{16}{9}, - 2, . . . . . .$
	Answered by Frix last updated on 17/Aug/23

	$\frac{16}{4}, - 2, \frac{16}{9}, - 2, \frac{16}{14}, - 2, \frac{16}{19}, . . .$ $\frac{16}{4}, - 2, \frac{16}{9}, - 2, \frac{16}{16}, - 2, \frac{16}{25}, . . .$ $I can find 42 answers per minute . . .$
	Commented by mathlove last updated on 17/Aug/23

	$s o l v e i t n t e r m$
	Commented by Frix last updated on 18/Aug/23

	$2 possible solutions out of {(1 zillion)}^{1 zillion}$ $a_{n} = - \frac{2}{9} (13 n^{3} - 100 n^{2} + 236 n - 167)$ $b_{n} = - \frac{32 n^{3}}{479 n^{3} - 3249 n^{2} + 6530 n - 3768}$
	Commented by Frix last updated on 18/Aug/23
	$...2 more c_n = { ((((64)/((n+3)^2 )); n=2k−1)),((−2; n=2k)) :}; k∈N^★ d_n = { ((((32)/(5n+3)); n=2k−1)),((−2; n=2k)) :}; k∈N^★$
	$. . . 2 more$ $c_{n} = {\begin{cases} \frac{64}{{(n + 3)}^{2}}; n = 2 k - 1 \\ - 2; n = 2 k \end{cases}; k \in N^{★}$ $d_{n} = {\begin{cases} \frac{32}{5 n + 3}; n = 2 k - 1 \\ - 2; n = 2 k \end{cases}; k \in N^{★}$
	Commented by Frix last updated on 18/Aug/23
	$... F_n =F_(n−2) +F_(n−1) ; F_1 =F_2 =1 e_n = { ((((64)/((n+3)^2 )); n=2k−1)),((−2F_(n/2) ; n=2k)) :}; k∈N^★ f_n = { ((((32)/(5n+3)); n=2k−1)),((−2F_(n/2) ; n=2k)) :}; k∈N^★$
	$. . .$ $F_{n} = F_{n - 2} + F_{n - 1}; F_{1} = F_{2} = 1$ $e_{n} = {\begin{cases} \frac{64}{{(n + 3)}^{2}}; n = 2 k - 1 \\ - 2 F_{\frac{n}{2}}; n = 2 k \end{cases}; k \in N^{★}$ $f_{n} = {\begin{cases} \frac{32}{5 n + 3}; n = 2 k - 1 \\ - 2 F_{\frac{n}{2}}; n = 2 k \end{cases}; k \in N^{★}$
	Commented by mathlove last updated on 18/Aug/23

	$t h a n k s$
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