# How can we prove this property?

$\qquad \qquad$ The greatest term in $\left(1+x\right)^{2n}$ has the greatest cofficient if $\frac{n}{n+1} \lt x \lt \frac{n+1}{n}$

Can we derive something like this for $n$ in general? Please give me some hints.

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I may be misunderstanding the question, but aren't the coefficients in the expansion independent of the value of $x$? – Alex Basson Dec 12 '10 at 16:33
You know that the middle terms are where the binomial coefficients are greatest, yes? – J. M. Dec 12 '10 at 16:34
@Alex Basson:Nopes consider of example that If $x$ lies in ($\frac{5}{6},\frac{6}{5}$) then the numerically greatest term in the expansion of $(1-x)^{21}$ has the numerically greatest coefficient. – Quixotic Dec 12 '10 at 16:37
@Alex : If $x=0.0001$, and $n=2$ then the greatest term is the first term, which is $1$, for instance. But it is the term in the binomial expansion with the smallest coefficient, because the middle term has coefficient $2$. – Raskolnikov Dec 12 '10 at 16:46
There is an article about the greatest term which might help (scroll down to the relevant section on the greatest term) tutors4you.com/binomialtheoremtutorial.htm – Timothy Wagner Dec 12 '10 at 16:49