find-the-general-term-of-the-sequence-3-5-9-17-33-

Question Number 159221 by physicstutes last updated on 14/Nov/21

$$\mathrm{find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{term}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{sequence}\: \\ $$$$\mathrm{3},\:\mathrm{5}\:,\:\mathrm{9},\:\mathrm{17},\:\mathrm{33},\:… \\ $$

Commented by MJS_new last updated on 14/Nov/21

a_n =(1/(12))n^4 −(1/2)n^3 +((23)/(12))n^2 −(3/2)n+3

$${a}_{{n}} =\frac{\mathrm{1}}{\mathrm{12}}{n}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{2}}{n}^{\mathrm{3}} +\frac{\mathrm{23}}{\mathrm{12}}{n}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}{n}+\mathrm{3} \\ $$

Commented by gsk2684 last updated on 14/Nov/21

$${pls}\:{explain}\:{ser} \\ $$

Commented by MJS_new last updated on 14/Nov/21

you can always find a polynome of degree k−1 for k given values. there′s never a unique solution for a question like this. why? because it′s possible to find at least one general term for any a_(k+1) ; in your example, I can give a solution for 3, 5, 9, 17, 33, a_6 for any a_6 ∈C

$$\mathrm{you}\:\mathrm{can}\:\mathrm{always}\:\mathrm{find}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree} \\ $$$${k}−\mathrm{1}\:\mathrm{for}\:{k}\:\mathrm{given}\:\mathrm{values}. \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{never}\:\mathrm{a}\:\mathrm{unique}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{a}\:\mathrm{question} \\ $$$$\mathrm{like}\:\mathrm{this}.\:\mathrm{why}?\:\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find} \\ $$$$\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{general}\:\mathrm{term}\:\mathrm{for}\:\mathrm{any}\:{a}_{{k}+\mathrm{1}} ;\:\mathrm{in} \\ $$$$\mathrm{your}\:\mathrm{example},\:\mathrm{I}\:\mathrm{can}\:\mathrm{give}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{for} \\ $$$$\mathrm{3},\:\mathrm{5},\:\mathrm{9},\:\mathrm{17},\:\mathrm{33},\:{a}_{\mathrm{6}} \:\mathrm{for}\:\mathrm{any}\:{a}_{\mathrm{6}} \in\mathbb{C} \\ $$

Commented by physicstutes last updated on 15/Nov/21

$$\mathrm{thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sirs} \\ $$

Answered by qaz last updated on 14/Nov/21

a_(n+1) −a_n =2^n ⇒Σ_(k=1) ^n (a_(k+1) −a_k )=Σ_(k=1) ^n 2^k ⇒a_(n+1) =((2(1−2^n ))/(1−2))+a_1 =2^(n+1) −2+3=2^(n+1) +1 ⇒a_n =2^n +1

$$\mathrm{a}_{\mathrm{n}+\mathrm{1}} −\mathrm{a}_{\mathrm{n}} =\mathrm{2}^{\mathrm{n}} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{a}_{\mathrm{k}+\mathrm{1}} −\mathrm{a}_{\mathrm{k}} \right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{2}^{\mathrm{k}} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}^{\mathrm{n}} \right)}{\mathrm{1}−\mathrm{2}}+\mathrm{a}_{\mathrm{1}} =\mathrm{2}^{\mathrm{n}+\mathrm{1}} −\mathrm{2}+\mathrm{3}=\mathrm{2}^{\mathrm{n}+\mathrm{1}} +\mathrm{1} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{n}} =\mathrm{2}^{\mathrm{n}} +\mathrm{1} \\ $$

Answered by floor(10²Eta[1]) last updated on 15/Nov/21

a_1 =a_1 a_2 =a_1 +2 a_3 =a_2 +2^2 a_4 =a_3 +2^3 ... a_n =a_(n−1) +2^(n−1) S_n =a_1 +(S_n −a_n )+(2+2^2 +...+2^(n−1) ) a_n =a_1 +2^n −2 a_n =2^n +1

$$\mathrm{a}_{\mathrm{1}} =\mathrm{a}_{\mathrm{1}} \\ $$$$\mathrm{a}_{\mathrm{2}} =\mathrm{a}_{\mathrm{1}} +\mathrm{2} \\ $$$$\mathrm{a}_{\mathrm{3}} =\mathrm{a}_{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{4}} =\mathrm{a}_{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} \\ $$$$… \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{a}_{\mathrm{n}−\mathrm{1}} +\mathrm{2}^{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{S}_{\mathrm{n}} =\mathrm{a}_{\mathrm{1}} +\left(\mathrm{S}_{\mathrm{n}} −\mathrm{a}_{\mathrm{n}} \right)+\left(\mathrm{2}+\mathrm{2}^{\mathrm{2}} +…+\mathrm{2}^{\mathrm{n}−\mathrm{1}} \right) \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{a}_{\mathrm{1}} +\mathrm{2}^{\mathrm{n}} −\mathrm{2} \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{2}^{\mathrm{n}} +\mathrm{1} \\ $$

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