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In my assignment I have the following question:

question

Alan

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Alan
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  • For some basic information about writing math at this site see e.g. here, here, here and here. – AlexR Mar 04 '15 at 21:40
  • Thank you. Is it bad I'm writing it the way I did? – Alan Mar 04 '15 at 21:41
  • The problem is that it's not searchable, doesn't show up in the question preview (front page "teaser" of the body) and looks pixelated. Also it consumes more bandwidth than necessary so, yes the format is bad. However, the question itself is asked nicely, so I'm convinced you'll learn MathJax and keep posting good questions like this one ;) – AlexR Mar 04 '15 at 21:43
  • Thank you! I will study the format, and I thank you for your tips! – Alan Mar 04 '15 at 21:44
  • Also note that you can inspect any Math $\mathfrak{such\ as\ this}$ with a right-click and "Show Math As $\blacktriangleright$ TeX commands". Try it out with my answer for example ;) – AlexR Mar 04 '15 at 21:45

3 Answers3

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Use a more straight-forward approach: $$4^n = (1+1)^{2n} = \sum_{i=0}^{2n} \binom{2n}i 1^i 1^{2n-i} = \sum_{i=0}^{2n} \binom{2n}i$$ Now since binomial coefficients are nonnegative, we get $$4^n \ge \binom{2n}i \qquad \forall 0\le i\le 2n$$ Especially for $i=n$.

AlexR
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In this answer, a simple argument shows that $$ \binom{2n}{n}\frac{\sqrt{n+\frac13}}{4^n} $$ is decreasing. Therefore, for any $n\ge0$, we have $$ \binom{2n}{n}\le\frac{4^n}{\sqrt{3n+1}} $$

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robjohn
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Alan.Here is my suggestion.
consider $(1+x)^{2n}$, then the coefficient of $x^n$ is $\binom{2n}{n}$.
we notice $(1+x)^{2n}$=$\sum_{i=0}^{2n}\binom{2n}{i}x^i$,and $\binom{2n}{i}\geq 1$.
So if we let x=1, then we have $\binom{2n}{n}\leq\sum_{i=0}^{2n}\binom{2n}{i}1^i=\sum_{i=0}^{2n}\binom{2n}{i}$.
and when x=1,$\sum_{i=0}^{2n}\binom{2n}{i}=(1+1)^{2n}=2^{2n}=4^n$.
Finally, we have $\binom{2n}{n} \leq 4^n$.

yakaqi
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