For a certain sequence of numbers, the sum of the first $n$ numbers in the sequence is given by $n^3+4n$ for all positive integers $n$. What is the fifteenth number in the sequence?
How do you solve this problem the most efficient way?
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4 Answers
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Solve[Table[Sum[f[i], {i, 1, n}] == n^3 + 4 n, {n, 15}], Array[f, 15]]
(*
{{f[1] -> 5, f[2] -> 11, f[3] -> 23, f[4] -> 41, f[5] -> 65,
f[6] -> 95, f[7] -> 131, f[8] -> 173, f[9] -> 221, f[10] -> 275,
f[11] -> 335, f[12] -> 401, f[13] -> 473, f[14] -> 551,
f[15] -> 635}}
*)
answered Oct 9, 2015 at 4:13
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$\begingroup$ "the most efficient way" is the one that takes you less time $\endgroup$ Oct 9, 2015 at 4:15
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$\begingroup$ I love this for not doing any analysis of the problem; just plugin what you know and have it solved. $\endgroup$ Oct 9, 2015 at 4:18
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$\begingroup$ @BrettChampion ha! That's why I wrote the comment above :) $\endgroup$ Oct 9, 2015 at 4:19
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$\begingroup$ My solutions were variants of
Differences[Table[n^3 + 4 n, {n, 0, 15}]], but that requires a step of reasoning first. $\endgroup$ Oct 9, 2015 at 4:20 -
1$\begingroup$ @BrettChampion Thinking is cheating :) $\endgroup$ Oct 9, 2015 at 4:20
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(FindSequenceFunction@ Differences[Prepend[n^3 + 4 n /. n -> Range@10, 0]])@15
(* 635 *)
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Might as well...
DifferenceDelta[n^3 + 4 n, n] /. n -> 15 - 1
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1$\begingroup$ That's the one function I always forget among the 848747 built-ins ;-} +1 $\endgroup$– ciaoOct 9, 2015 at 6:03
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$\begingroup$ Familiarity with the difference calculus has so far served me well. :) $\endgroup$ Oct 9, 2015 at 6:08
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Let $S_n=\sum^n_{j=1}a_n$. So $a_n=S_n-S_{n-1}$.
5 - 3 n + 3 n^2 /. n -> 15
yields 635. (Mathematica can obviously do this).
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$\begingroup$ I saw the beginning of your answer and thought you were going to give an answer suitable for math.se. $\endgroup$ Oct 9, 2015 at 6:41
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$\begingroup$ @BrettChampion...I can delete if you feel this is not in the spirit of the site. I just thought given the precise expression of the question, this was the simplest approach... $\endgroup$– ubpdqnOct 9, 2015 at 6:44
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$\begingroup$ This is more or less what
DifferenceDelta[]does, made explicit. $\endgroup$ Oct 9, 2015 at 8:15 -
$\begingroup$ @J. M. yes...I am sorry if it seemed like being a wise-guy...I just thought simplicity was warranted...I did upvote your answer as I was not aware of
DifferenceDelta. I am happy to delete and acknowledge all the answers get to the same result. :) $\endgroup$– ubpdqnOct 9, 2015 at 8:19 -
$\begingroup$ No, I didn't take it like that! I was merely adding an explanatory note; at least, OP will need to think a little bit more for your solution. :) $\endgroup$ Oct 9, 2015 at 8:21
s(15)-s(14)$\endgroup$