# How could we find the largest number in the sequence $\sqrt{50},2\sqrt{49},3\sqrt{48},\cdots 49\sqrt{2},50$?

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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If you square all of them, the largest of the squares will correspond to the largest of the square roots. –  Ｊ. Ｍ. Nov 8 '11 at 10:45
@GerryMyerson Shame on me :) I did not read the question carefully enough. –  Sasha Nov 8 '11 at 13:42

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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+1 Well spotted! –  Jyrki Lahtonen Nov 8 '11 at 13:19
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. –  pharmine Nov 9 '11 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.

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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!" –  Jyrki Lahtonen Nov 8 '11 at 14:21
Nice solution (+1). I have just use a "calculus approach". –  Tapu Nov 8 '11 at 14:48
Well, you used the difference calculus, but so what? This is neat! –  Ｊ. Ｍ. Nov 8 '11 at 14:59
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,...
Sorry the point $x=0$ is not within the domain, so need not be considered. –  Tapu Nov 8 '11 at 14:43
+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 :-) –  Jyrki Lahtonen Nov 8 '11 at 20:14