How to find the largest number in the sequence$$ \sqrt{50},2\sqrt{49},3\sqrt{48},\cdots 49\sqrt{2},50$$

I am interested in a "calculus-free" approach. Thanks,

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    $\begingroup$ If you square all of them, the largest of the squares will correspond to the largest of the square roots. $\endgroup$ Nov 8, 2011 at 10:45
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    $\begingroup$ @GerryMyerson Shame on me :) I did not read the question carefully enough. $\endgroup$
    – Sasha
    Nov 8, 2011 at 13:42

3 Answers 3


The $n$-th term in the sequence is $n\sqrt{51-n}=\sqrt{n^2(51-n)}$. So the question is: for which $n$ ($1\le n\le 50$), does $n^2(51-n)$ become the largest?

If you want to avoid calculus, you could use the AM-GM inequality: if $x,\,y,\,z\ge 0$, then $$\frac{x+y+z}{3}\ge\sqrt[3]{xyz},$$ with equality if and only if $x=y=z$.

If we set $x=y=n/2$ and $z=51-n$, we obtain: $$\frac{51}{3}\ge \sqrt[3]{\frac{n}{2}\cdot\frac{n}{2}\cdot (51-n)},$$ with equality if and only if $n/2=51-n$ or $n=34$.

It follows that $n^2(51-n)\le 4\cdot 17^3$, or $\sqrt{n^2(51-n)}\le 2\cdot 17^{3/2}$, where equality holds for $n=34$.

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    $\begingroup$ Note that if the original question is for example $n\sqrt{50-n}$ instead of $n\sqrt{51-n}$, no integer $n$ satisfies the equality of the AM-GM inequality; in that case, a more careful evaluation, such as that used in Jyrki Lahtonen's answer, is needed. $\endgroup$
    – pharmine
    Nov 9, 2011 at 4:39

I like the AM-GM argument a lot, but here's another more down-to-earth solution.

Write $a_n=n\sqrt{51-n}$, $1\le n\le 50$. Everything in sight is positive, so $a_n\le a_{n+1}$ if and only if $$ a_n^2\le a_{n+1}^2\Leftrightarrow n^2(51-n)\le (n+1)^2(50-n). $$ This latter inequality simplifies to the quadratic inequality $-3n^2+99n+50\ge0$. The plot of this function is a parabola opening downwards. Therefore the inequality holds between the zeros $n_1\approx-0.5$ and $n_2\approx 33.5$.

We have shown that $a_{n+1}$ is larger than $a_n$, when $1\le n\le 33$, and that $a_{n+1}$ is smaller than $a_n$, when $n\ge 34$. Therefore we can conclude that $a_{34}$ is the largest of this lot.

  • $\begingroup$ On second thought: I used $f(n+1)-f(n)$ as a crude substitute for $\lim_{h\to0}(f(n+h)-f(n))/h$, so it's a bit questionable, whether this really is a non-calculus approach? Well, Abhyankar once said in a plenary talk that he uses $[f(x+y)-f(x)]$ instead of a derivative. "Because it always exists!" $\endgroup$ Nov 8, 2011 at 14:21
  • $\begingroup$ Nice solution (+1). I have just use a "calculus approach". $\endgroup$
    – Tapu
    Nov 8, 2011 at 14:48
  • $\begingroup$ Well, you used the difference calculus, but so what? This is neat! $\endgroup$ Nov 8, 2011 at 14:59

I would like to apply calculus whenever it is possible. So, here is my trying (IMO this is not a better solution though):

Consider the function $f(x)=x^2(51-x)$ over $[1,50]$. Then as usual, $f'(x)=0\Rightarrow x=0,34$ and $f''(0)>0,f''(34)<0$ implies $f$ has a unique global maximum at $x=34$ and global minimum at $x=0$. So,...

  • $\begingroup$ Sorry the point $x=0$ is not within the domain, so need not be considered. $\endgroup$
    – Tapu
    Nov 8, 2011 at 14:43
  • $\begingroup$ +1 for the effort. If you view this as a problem of a continuous variable $x$ as opposed to the described discrete variable $n$, then you might as well extend the range to $[0,51]$. I mean, then it is easier to evaluate your function at the end points, and with this type of "find the extremum" problem the answer is unlikely to be at the endpoint anyway :-) $\endgroup$ Nov 8, 2011 at 20:14
  • $\begingroup$ @JyrkiLahtonen: I did considered this problem as a continuous problem. Whats wrong with that? The given set of points is indeed a finite subset of the set of points I am considering. The point of maximum of this superset also belongs to the subset under consideration. So it should be the maximum. Is there anything I am missing? $\endgroup$
    – Tapu
    Nov 8, 2011 at 20:40
  • $\begingroup$ No. Nothing wrong with your solution. But, if the zero of the derivative had not occurred at an integer point you would have had to work a bit more :-) $\endgroup$ Nov 8, 2011 at 20:43
  • $\begingroup$ Yes, thats why I wrote "IMO its not a better solution". However, I wish to mention that a particular problem can be tackled in a tricky way which may not be applicable to general problem. For this problem, may be I was lucky that it happens (indeed I was lucky today: if you don't mind, here is another example). I wish to thank you for having the patience of verifying and clarifying (yes some people just voted up/down probably without verifying). $\endgroup$
    – Tapu
    Nov 8, 2011 at 21:14

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