Help with Binomial Theorem: Greatest coefficient

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.

• 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? – Rory Daulton Feb 18 '15 at 11:39
• the entire term ${n \choose k}a^{n-k}b^k$ – Justin HT Feb 18 '15 at 11:41
• @RoryDaulton The signed value; and yeah, you're right about the word, but I was sort of quoting my textbook... – Justin HT Feb 18 '15 at 11:44

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.

• 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) – Justin HT Feb 18 '15 at 12:03

the relation works if you make certain changes, numerically greatest term means greatest absolute value so $$|T_{k+1}|>|T_k|$$ thus for an expansion $$(a+bx)^n$$ we have $$\frac{n-k+1}{k} \cdot \mid\frac{bx}{a}\mid \geq 1$$ $$\implies k \leq \frac{n+1}{1+\mid\frac{a}{bx}\mid}$$