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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.

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  • $\begingroup$ What do you mean by "the greatest coefficient in a binomial expansion"? Do you mean the coefficient that goes in front of $a^{n-k}b^k$, namely ${n \choose k}$, or do you mean the entire term ${n \choose k}a^{n-k}b^k$? The word "coefficient" usually means the former, but your work seems to mean the latter. And by "greatest" do you mean in signed value or in absolute value? $\endgroup$ – Rory Daulton Feb 18 '15 at 11:39
  • $\begingroup$ the entire term ${n \choose k}a^{n-k}b^k$ $\endgroup$ – Justin HT Feb 18 '15 at 11:41
  • $\begingroup$ @RoryDaulton The signed value; and yeah, you're right about the word, but I was sort of quoting my textbook... $\endgroup$ – Justin HT Feb 18 '15 at 11:44
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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.

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  • $\begingroup$ Thanks kryomaxim for answering but I'm not sure how to apply to a question: e.g. $(4*x -3*y)^9 $ I solve this and I got $ k \leq -30 $ (I verified with Wolfram Alpha here "wolframalpha.com/input/?i=-%2810-k%29%2Fk+*%283%2F4%29+%5Cleq+1" ) I'm not sure if this is correct, (also if it's yes how to interpret this result) $\endgroup$ – Justin HT Feb 18 '15 at 12:03

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