So I have this list: $a_0=1, a_1=15, a_2=150, a_3=1250, a_4=9375, a_5=65625, a_6=437500$. I need to find a formula for $a_i$ and prove that $a_{20} = 22029876708984375$. I've found that formula for $a_i = \frac {5^{i-1}\cdot i\cdot (i+1)}{2}$. But I have no idea how I should have had to come up with this formula. What is the way to show how I had came up with this formula(I need to explain to teacher it step by step)?
How to find formula for i element
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calculus
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0"Find the next term". Given enough computational power I can make $a_{20}$ be any number at all and justify it with a polynomial (see Lagrange interpolation). What restrictions must there be on the general formula? – 2017-01-13
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0Since there is not a unique formula that fits the five points, there is no systematic way to arrive at such a formula. – 2017-01-13
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2These things are usually matters for guesswork....your list is very brief. I assume, by the way, that you meant $a_n=5^{n-1}\frac {n(n+1)}2$. You could guess that by extracting powers of $5$ and recognizing the triangular numbers as the quotient. – 2017-01-13
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0Just to say, the interpolating polynomial isn't very pleasant: $\frac {652}3 x^4 - \frac {3490}3 x^3 + \frac {12175}6x^2 - \frac{6415}6 x + 1$ Seems unlikely that this is what was intended. – 2017-01-13
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0I updated the formula, which I wrote wrong at first. – 2017-01-13
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0@lulu I'm a little bit confused, since english isn't my native language and math isn't my best friend, but could You please explain it in a little bit other words? – 2017-01-13
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0No problem. The difficulty here is that your list is very short. Lots of sequences match these terms so you have to guess what the writer intended. In this case, someone might notice that $5$ divides each term a lot. Indeed, $5^{n-1}$ divides $a_n$. If you remove that factor you have $1,3,6,10,15$. Those are the triangular numbers! That is $T(n)=1+2+\cdots+n=\frac {n(n+1)}2$. Very likely guess. Or you might look for a polynomial that interpolates these numbers...in a prior comment I gave a polynomial $p(x)$ and you can check that $a_n=p(n-1)$ (I started the interpolation at $0$). ... – 2017-01-13
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0@lulu Unfortunately, $5^{20-1}\cdot 20\cdot21/2$ is way short of the given value of $a_{20}$. – 2017-01-13
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0...So the coefficients could be derived from that polynomial. But it's an ugly polynomial, so I doubt this is what was intended. But there's more Art here than Science. – 2017-01-13
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0@egreg Yeah, the specified number is $a_{21}$. Seems like a fencepost error of some sort. – 2017-01-13
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0For what it's worth; I got the triangular number answer by inspection (well, with simple calculations). I got the polynomial with the help of a computer. Had the coefficients of the polynomial turned out to be simple integers, I'd have suspected that this was the intended answer. As it is...well, I doubt anybody wanted you to go down that route. – 2017-01-13
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0@lulu My preferred sequence is $5,\;10,\;20,\;30,\;36,\;?$ (guess the next one) – 2017-01-13
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0@egreg Ha! Lucky for me I know my Mozart. Well, turns out I don't know it all that well...I still had to look up the next one. – 2017-01-13
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0@lulu I have the advantage I'm a native Italian speaker, so I know what number goes with the verses. `;-)` – 2017-01-13
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0@lulu I'm pretty sure I understood that now. Thanks a lot, You da real MVP :) – 2017-01-13
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0@lulu wait, one more thing: I understood eveerything you wrote. But how can I come from "all those numbers can be divided by 5" to "$5^{i-1}$"? The thing is that I must explain every step. – 2017-01-13
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0You just edited your post and changed the numbering system That makes all the prior comments slightly off. Now the correct formula is $a_n=5^n\times T(n+1)$.Thus $a_0=5^0\times T(1)=1$ and so on. – 2017-01-13
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0@lulu Yeah I changed th epost, because in my notes everything is messed up, but it's the final version. But, again, I know what triangle numbers are. But how can I explain how did I come up with those triangle numbers. Because at start we can suggest that all those numbers(15, 150, 1250) can be divided by 5. But there is no more further connection between 5s and triangle numbers. E.g. how did you come up with this new formula – 2017-01-13
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0Again, we see that the power of $5$ which divides each $a_n$ grows with $n$. We seek a pattern to that growth, and it's not too big a stretch to see that (with your new numbering) $5^n\,|\,a^n$. We then ask what $\frac {a_n}{5^n}$ looks like and, writing them down, we see the triangular numbers. I don't think it gets more formal than that...as I say, there are polynomials that work too (the one I wrote down no longer works, I expect, because you added new terms. But there will be a higher degree polynomial that works. – 2017-01-13
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0@lulu yeah right, it's really easy. I've just got really messed up. Thanks again for your help) – 2017-01-13
2 Answers
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One can "guess" that the sequence if the $5$-th binomial transform of $(0,0,1,0,0,0,.......)$, see OEIS. Then $$ a_n=5^{n-2}\binom{n}{2}, $$ with a shift of $2$ (so then $a_{22}=22029876708984375$). Mathematical "guessing" from given numbers has a precise meaning, and is used in algebraic combinatorics and other fields.
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0thanks for help, but in my case I have to show how I came up with formula step by step, without anything which was already done. I hope I explained it right – 2017-01-13
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0You are welcome, but in your case you cannot "show" how you came up with the formula step by step, because there is more than one answer, and you could come up with some different formula, too. One can only explain how "guessing algorithms" will most likely discover this formula. – 2017-01-13
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ADDED: I suppose the first thing I would do is say that there seemed to be a power of $5$ involved, and consider $$ \frac{a_n}{5^n} = 1,3,6, 10, 15, 21,... $$ which are the triangular numbers
What you want is some of the extended methods in The Book of Numbers by Conway and Guy. Since your sequence is a mixture, the main "number walls" technique may not be enough; you may need to look up the "number fans" method. The point is, though, they consider a specific set of mathematical rules for creating a sequence, and show ways to figure that out. No guess work.
