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

faind n terme 4,−2,((16)/9),−2,......

$${faind}\:{n}\:{terme} \\ $$$$\mathrm{4},−\mathrm{2},\frac{\mathrm{16}}{\mathrm{9}},−\mathrm{2},...... \\ $$

Answered by Frix last updated on 17/Aug/23

((16)/4),−2,((16)/9),−2,((16)/(14)),−2,((16)/(19)),... ((16)/4),−2,((16)/9),−2,((16)/(16)),−2,((16)/(25)),... I can find 42 answers per minute...

$$\frac{\mathrm{16}}{\mathrm{4}},−\mathrm{2},\frac{\mathrm{16}}{\mathrm{9}},−\mathrm{2},\frac{\mathrm{16}}{\mathrm{14}},−\mathrm{2},\frac{\mathrm{16}}{\mathrm{19}},... \\ $$$$\frac{\mathrm{16}}{\mathrm{4}},−\mathrm{2},\frac{\mathrm{16}}{\mathrm{9}},−\mathrm{2},\frac{\mathrm{16}}{\mathrm{16}},−\mathrm{2},\frac{\mathrm{16}}{\mathrm{25}},... \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{find}\:\mathrm{42}\:\mathrm{answers}\:\mathrm{per}\:\mathrm{minute}... \\ $$

Commented by mathlove last updated on 17/Aug/23

solve it n term

$${solve}\:{it}\:{n}\:{term} \\ $$

Commented by Frix last updated on 18/Aug/23

2 possible solutions out of (1zillion)^(1zillion) a_n =−(2/9)(13n^3 −100n^2 +236n−167) b_n =−((32n^3 )/(479n^3 −3249n^2 +6530n−3768))

$$\mathrm{2}\:\mathrm{possible}\:\mathrm{solutions}\:\mathrm{out}\:\mathrm{of}\:\left(\mathrm{1zillion}\right)^{\mathrm{1zillion}} \\ $$$${a}_{{n}} =−\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{13}{n}^{\mathrm{3}} −\mathrm{100}{n}^{\mathrm{2}} +\mathrm{236}{n}−\mathrm{167}\right) \\ $$$${b}_{{n}} =−\frac{\mathrm{32}{n}^{\mathrm{3}} }{\mathrm{479}{n}^{\mathrm{3}} −\mathrm{3249}{n}^{\mathrm{2}} +\mathrm{6530}{n}−\mathrm{3768}} \\ $$

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^★

$$...\mathrm{2}\:\mathrm{more} \\ $$$${c}_{{n}} =\begin{cases}{\frac{\mathrm{64}}{\left({n}+\mathrm{3}\right)^{\mathrm{2}} };\:{n}=\mathrm{2}{k}−\mathrm{1}}\\{−\mathrm{2};\:{n}=\mathrm{2}{k}}\end{cases};\:{k}\in\mathbb{N}^{\bigstar} \\ $$$${d}_{{n}} =\begin{cases}{\frac{\mathrm{32}}{\mathrm{5}{n}+\mathrm{3}};\:{n}=\mathrm{2}{k}−\mathrm{1}}\\{−\mathrm{2};\:{n}=\mathrm{2}{k}}\end{cases};\:{k}\in\mathbb{N}^{\bigstar} \\ $$

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^★

$$... \\ $$$$\mathcal{F}_{{n}} =\mathcal{F}_{{n}−\mathrm{2}} +\mathcal{F}_{{n}−\mathrm{1}} ;\:\mathcal{F}_{\mathrm{1}} =\mathcal{F}_{\mathrm{2}} =\mathrm{1} \\ $$$${e}_{{n}} =\begin{cases}{\frac{\mathrm{64}}{\left({n}+\mathrm{3}\right)^{\mathrm{2}} };\:{n}=\mathrm{2}{k}−\mathrm{1}}\\{−\mathrm{2}\mathcal{F}_{\frac{{n}}{\mathrm{2}}} ;\:{n}=\mathrm{2}{k}}\end{cases};\:{k}\in\mathbb{N}^{\bigstar} \\ $$$${f}_{{n}} =\begin{cases}{\frac{\mathrm{32}}{\mathrm{5}{n}+\mathrm{3}};\:{n}=\mathrm{2}{k}−\mathrm{1}}\\{−\mathrm{2}\mathcal{F}_{\frac{{n}}{\mathrm{2}}} ;\:{n}=\mathrm{2}{k}}\end{cases};\:{k}\in\mathbb{N}^{\bigstar} \\ $$

Commented by mathlove last updated on 18/Aug/23

thanks

$${thanks} \\ $$

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