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Find a general formula for the terms of the sequence

$${a_n}=\left\{ \frac{11}{7},\frac{107}{49},\frac{659}{343},\frac{4883}{2401},\frac{33371}{16807},\frac{234569}{117649},\dots \right\}$$

I don't know how to approach this question as it is not arithmetic or geometric. I know the denominator is geometric increasing by a factor of $7$ but I can't find what the numerator should be for the general formula for the terms. Anyone know what it is?

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    $\begingroup$ Where does it come from? $\endgroup$ – Jean-Claude Arbaut Apr 21 at 17:14
  • $\begingroup$ My online calculus questions platform called Mobius(used to be Maple TA). I believe it must be a mistake as nothing I input is giving me the correct answer. $\endgroup$ – RaV1oLLi Apr 21 at 17:16
  • $\begingroup$ Have you put it into OEIS? $\endgroup$ – Dave Apr 21 at 17:22
  • $\begingroup$ This is in the OEIS, but not much else - oeis.org/A081657 $\endgroup$ – Peter Foreman Apr 21 at 17:23
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    $\begingroup$ Nobody can tell you the correct answer. Whatever answer you give, the proposer can tell you that you are wrong. You an't win this kind of game. $\endgroup$ – Somos Apr 21 at 18:55
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The general formula for an (infinite) sequence of (e. g. real) numbers from the finite number $n$ of its (first) members is in principle impossible, as the next (not listed) $(n+1)^\mathrm{th}$ member may be an arbitrary number, and there is still a formula for expressing $a_1, \dots, a_n, a_{n+1},$ e. g. as a polynomial of order $n$:

$$a_k = \sum_{i=0}^nb_ik^i,\quad k = 1, \dots,n+1$$

The process for finding coefficients $b_0, \dots, b_n$ is straightforward enough.

In other words, if someone will find the formula for your "sequence", there is still the infinity number of other formulas, giving different sequences, but all of them producing your "sequence", i. e. $$\frac{11}{7},\frac{107}{49},\frac{659}{343},\frac{4883}{2401},\frac{33371}{16807},\frac{234569}{117649}.$$


Note:

It means that all psychological tests of type

What is the next number of the sequence $1, 2, 3, 4, 5?$

are in principle meaningless ones, because you may tell "$1762$", and then show to surprised psychologist a formula supporting the correctness of your answer:

$$a_k = {439\over30}k^5-{439\over 2}k^4+{7463\over 6}k^3-{6585\over 2}k^2+{60158\over 15}k-1756$$

If he/she will not trust you, launch SageMath, which will produce accurate, non-rounded results, and write commands

sage: var("k")
sage: a(k) = (439/30)*k^5 - (439/2)*k^4+(7463/6)*k^3-(6585/2)*k^2+(60158/15)*k-1756
sage: a(1), a(2), a(3), a(4), a(5), a(6), a(7), a(8)

to obtain the result

(1, 2, 3, 4, 5, 1762, 10543, 36884)

(and to give the psychologist two more members for free).


Note 2:

It doesn't mean that there is not a simpler formula - including a recurrent one or other "recipe" - for the same (finite) sequence.

For example, there is so simple one for the rather not so trivial sequence

$$\color{blue}{1, 11, 21, 1211, 111221}$$

that even 6-7 year-old child is able to write down the next element ($\color{red}{312211}$) if you tell it the rule, or - perhaps - even without telling it.

No, you have no chance to discover this simple rule (supposing your age is $10$+). Don't waste your time. It's a good advice, believe me.

(Googling for it is a much better approach.)

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The general term seems to be $$a_n=\frac{2(7^n)+(-3)^n}{7^n}=2+\left(-\frac{3}{7}\right)^n$$ But the last term is given by $$a_6=2-\left(\frac{3}{7}\right)^6$$ so this formula does not always work. A suitable formula could be $$a_n=\begin{cases}2-\left(\frac{3}{7}\right)^n&n\equiv0\mod{6}\\ 2+\left(-\frac{3}{7}\right)^n &\text{otherwise} \end{cases}$$

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The numerator of the $n$-th term seems to be $$2\cdot 7^n+(-1)^n 3^n. $$

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  • $\begingroup$ May I ask how you got to this answer? Or is trial and error the only method?Edit: I inputted this answer into the online question and it came out as incorrect. $\endgroup$ – RaV1oLLi Apr 21 at 17:07
  • $\begingroup$ I was thinking the same, but what about the last term? $234569\ne 2(7^6)+3^6$ $\endgroup$ – Peter Foreman Apr 21 at 17:07
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    $\begingroup$ @PeterForeman Arg, yes, the last is $2(7)^6\color{red}{-}3^6$! $\endgroup$ – Jean-Claude Arbaut Apr 21 at 17:10
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    $\begingroup$ @RaV1oLLi: I noticed the numerators are the successive powers of $7$, and the numerators were not far from twice the denominator, with a corrective term. So I subtracted twice the denominators from the numerators, and tried to see a pattern. I tested only the 5 first terms. $\endgroup$ – Bernard Apr 21 at 17:20

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