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Question Number 176293 by mokys last updated on 15/Sep/22

find n^(th) formolla of <0,4,9,16,25>

$${find}\:{n}^{{th}} \:{formolla}\:{of}\:<\mathrm{0},\mathrm{4},\mathrm{9},\mathrm{16},\mathrm{25}> \\ $$

Answered by Frix last updated on 16/Sep/22

for <0, 4, 9, 16, 25> we get the polynomial a_n =−(((n−1)(n^3 −13n^2 +34n−120))/(24))

$$\mathrm{for}\:<\mathrm{0},\:\mathrm{4},\:\mathrm{9},\:\mathrm{16},\:\mathrm{25}>\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{polynomial} \\ $$ $${a}_{{n}} =−\frac{\left({n}−\mathrm{1}\right)\left({n}^{\mathrm{3}} −\mathrm{13}{n}^{\mathrm{2}} +\mathrm{34}{n}−\mathrm{120}\right)}{\mathrm{24}} \\ $$

Answered by Rasheed.Sindhi last updated on 16/Sep/22

p(n)= { ((n^2 −1 if n=1)),((n^2 otherwise)) :} n∈{1,2,3,...}

$${p}\left({n}\right)=\begin{cases}{{n}^{\mathrm{2}} −\mathrm{1}\:{if}\:{n}=\mathrm{1}}\\{{n}^{\mathrm{2}} \:{otherwise}}\end{cases}\:\:{n}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},...\right\} \\ $$

Answered by Rasheed.Sindhi last updated on 16/Sep/22

★ determinant (((A Simple Formula_(of nth term_(FOR THE SEQUENCE_(0,4,9,16,25,...) ) ) )))★ determinant ((( determinant (((a_n =n^2 −⌊(1/n)⌋ ; n∈{1,2,3,...})))_ ^ )))

$$\:\:\:\:\:\:\bigstar\begin{array}{|c|}{\underset{\underset{\underset{\mathrm{0},\mathrm{4},\mathrm{9},\mathrm{16},\mathrm{25},...} {\boldsymbol{\mathcal{F}}\mathcal{OR}\:\mathcal{THE}\:\mathcal{SEQUENCE}}} {\boldsymbol{{of}}\:\boldsymbol{{nth}}\:\boldsymbol{{term}}}} {\mathbb{A}\:\mathbb{S}\boldsymbol{\mathrm{imple}}\:\mathbb{F}\boldsymbol{\mathrm{ormula}}}}\\\hline\end{array}\bigstar \\ $$ $$\begin{array}{|c|}{\:\:\begin{array}{|c|}{{a}_{{n}} ={n}^{\mathrm{2}} −\lfloor\frac{\mathrm{1}}{{n}}\rfloor\:\:;\:{n}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},...\right\}}\\\hline\end{array}_{} ^{} }\\\hline\end{array} \\ $$

Commented bymr W last updated on 16/Sep/22

or a_n =sign(n−1)n^2

$${or} \\ $$ $${a}_{{n}} ={sign}\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} \\ $$

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