my textbook said to determine the greatest coefficient in a binomial expansion $ (a+b)^n $ we can use the inequality:
\begin{align} \frac{n-k+1}{k} \cdot \frac{b}{a} \geq 1 \end{align}
Then solve for $k$ which will result in something like $ k \leq constant$ which we can then substitute back to solve for the greatest coefficient.
my question is what about $(a-b)^n $? (where $a > 0$, $ b > 0$)
Solving for k would result in:
\begin{align} k \leq -\frac{b \cdot(n+1)}{a+1} \end{align}
Since $b$, $n$ and $a$ are all positive. Then $k$ appear to be a negative number (which is wrong)
Can someone tell me where did I got wrong? Thank you.
For $a \rightarrow a, b \rightarrow -b$ you get
$\frac{-(n-k+1)b}{ka} \geq 1,a>0,b>0$
$-(n-k+1)b \geq ka$ (if $k>0$!)
$-(n+1)b \geq k(a-b)$
$\frac{(n+1)b}{b-a}\geq k$ (only if $a>b$)
If $a<b$ the relation sign inverts! You get $k \geq \frac{(n+1)b}{b-a}$ in this case.