# 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.

Paulo Argolo

Rio de Janeiro, Brazil

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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 $k<\frac{|b|(n+1)}{(|b|+|a|)}-1$, so $f(k)$ is increasing until $k=I\left(\frac{|b|(n+1)}{(|b|+|a|)}\right)$ where $I$ denotes the integral part.
Maybe the notation $C^k_n$ is confusing you, it is also written $\binom{k}{n}$ and equal to $\frac{n!}{k!(n-k)!}$. To better understand the proof you can also do the case $a=b$, where only the binomial coefficients "count". –  Plop Nov 5 '10 at 1:21
$\binom{n}{k}$, not $\binom{k}{n}$. –  Hans Lundmark Nov 5 '10 at 7:50