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$ \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) – Timothy Wagner Dec 12 '10 at 16:49

Hint: I think you can solve this by looking at the adjacent terms of the middle term. If the middle term is the biggest term, it has to be bigger than both adjacent terms. This will give you the bounds.

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