Maximum term of (a + b) ^ n I would like a demonstration of the fact below.

Being given real numbers a and b (nonzero) and a positive integer n, the order p, that occupies the maximum term (in absolute value)
 of the development of power (a+b)^ n, according to decreasing powers of a is given by:

p = 1 + integer part of [|b|(n+1)/(|a|+|b|)]

When n is integer, there are maximum two terms: those of order p and p-1.

|a| and |b| are the modules of the numbers a and b, respectively.

Already, very grateful.

Paulo Argolo

Rio de Janeiro, Brazil
Let $f(k)=C_n^k |a|^{n-k} |b|^k$.

Then $f(k+1)/f(k)=\frac{|b|}{|a|} \frac{n-k}{k+1} > 1$ iff $k&lt;\frac{|b|(n+1)}{(|b|+|a|)}-1$, so $f(k)$ is increasing until $k=I\left(\frac{|b|(n+1)}{(|b|+|a|)}\right)$ where $I$ denotes the integral part.
Every positive integer is of the form 3k-2,3k-1, or 3k.  


If n is of the form 3k-2, then ak has the maximum value
If n is of the form 3k-1, then ak = a[k+1] have the maximum value.
If n is of the form 3k, then a[k+1] has the maximum value.


It's not hard to prove each of those.  Write out the terms
with the factorials and powers of 2 for positive j&lt;=k and prove: 
a[k-j] &lt; ak {and/or a[k+1]}  and ak {and/or a[k+1]} > a[k+1+j]  

Edwin McCravy