# Binomial series expansion

1. Jan 19, 2014

### sooyong94

1. The problem statement, all variables and given/known data
For $n>0$, the expansion of $(1+mx)^{-n}$ in ascending powers of $x$ is $1+8x+48x^{2}+...$

(a) Find the constants $m$ and $n$
(b) Show that the coefficient of $x^{400}$ is in the form of $a(4)^{k}$, where $a$ and $k$ are real constants.

2. Relevant equations
Binomial expansion

3. The attempt at a solution
I have expanded the series using binomial theorem, and compared each coefficient. Then I solved them using simultaneous equation. That works out to be $m=-4, n=2$.

(b) I know the binomial coefficient is $\frac{n(n-1)(n-2)...(n-r+1)}{r!}$, though I'm stuck on that... :(

Last edited: Jan 19, 2014
2. Jan 19, 2014

### Simon Bridge

if $n=2$, then how can there be an $x^{400}$ term? Why does the provided expansion not terminate?
if m=-4, as well, then wouldn't that mean that the second term in the expansion is -8x instead of the +8x given?

Last edited: Jan 19, 2014
3. Jan 19, 2014

### sooyong94

Oops... It should be $(1+mx)^{-n}$... Sorry about that. :P

4. Jan 19, 2014

### Simon Bridge

OK - so you have the expansion, in principle, of form:
$$(1+mx)^{-n}=\sum_{i=1}^N a_ix^i$$ ... the coefficient of $x^{400}$ is $a_{400}$ .... and you know that $a_i$ depends on n,m and r from the binomial expansion formula.

You have a formula for the binomial coefficient - but $a_{400}$ has an extra term due to the value of m.
The trick is to match up the values you have with the formulas ... i.e. while n=2 in the equation above, the "n" in the formula for the binomial coefficient is different ... it may help to rewrite the formula taking this into account.
If $a_{400}$ corresponds to the r'th binomial coefficient then what is r?

5. Jan 19, 2014

### sooyong94

I don't get it... :/
I know the coefficient of $x^{400}$ would look something like this:
$\frac{-2(-3)(-4)....(-400-2+1)}{400!}$
Though it doesn't look like what the question needs... :S

6. Jan 20, 2014

### Simon Bridge

You forgot about the value of m ;)

Anyway - you have to show that the series is the same as what they gave you.
Find the value of a and k.

Just to get you thinking the right way: is the product a positive or negative number?
Maybe it is -2x3x4x...x401 or +2x3x4x...x401

.... can you represent it in some kind of shorthand?
Say using factorial notation?

7. Jan 20, 2014

### sooyong94

Where's the value of m?

That actually looks like (401!) ?

8. Jan 20, 2014

### Simon Bridge

You said that m=-4 remember?

Notice that the question does not ask for the binomial coefficient.
The coefficient of x^400 includes the binomial coefficient but is not equal to it... just like the coefficient of x^2 is not equal to it's binomial coefficient. Compare.

That's right! So to ratio comes to 401!/400! doesn't it? ... what is that equal to?

9. Jan 20, 2014

### sooyong94

Still don't get it on the binomial coefficient though...

I know that 401!/400!=(401x400!)/(400!)=401

10. Jan 20, 2014

### Simon Bridge

That's all there is to it! Well done ;)
How did you work out that it should be positive rather than negative?

You need the part where "m=4" fits in to make the connection.
Remember, you need to show that $a_{400}=b(4)^k$ where #a_n$is the coefficient of$x^n$in the polynomial, with b and k both positive real numbers. 11. Jan 20, 2014 ### sooyong94 I don't know about that though... :/ But how do I use that m=4? 12. Jan 21, 2014 ### Simon Bridge Well consider: what is the coefficient of x and what is the coefficient of x^2? How does that m=-4 contribute to those coefficients? How does the binomial coefficient contribute? You calculated it - you should know! Write ut the coefficient of x and x^2 in terms of n and m again - instead of actual numbers. 13. Jan 21, 2014 ### sooyong94 Coefficient of x would look something like this: -nm Coefficient of x^2 would look something like this:$\frac{(-n)(-n-1))}{2!} m^{2}$14. Jan 22, 2014 ### haruspex Yes. m is just a multiplier on x, so you can start with the expansion for m=1 and just replace x with (mx) in the expansion. 15. Jan 22, 2014 ### sooyong94 So that means... the coefficient of$x^{400}$is$\frac{(-2)(-3)(-4)...(-400-2+1)}{400!} (-4)^{400}## ?

16. Jan 22, 2014

### haruspex

Yes, but you can get rid of all of those awkward minus signs. Each extra factor like (-400-2+1) inverts the sign, but so does the extra -4 term, so they all cancel.

17. Jan 22, 2014

### sooyong94

Erm... why? How does it invert the minus sign?
I know there are 400 terms, so the signs cancel each other, because 400 is an even number?

18. Jan 22, 2014

### haruspex

All the terms in the expansion are positive. (−2)(−3)(−4)...(−n−2+1) has n terms, so has the same sign as (-4)n.

19. Jan 22, 2014

### sooyong94

How does all terms in the expansion are positive?

20. Jan 22, 2014

### haruspex

The first term (1) is positive. To get the second term you multiply by (-2)(-4) = 8. To get the next you multiply be (-3)(-4) = 12. At every step you multiply by two factors, both negative, so the minus signs cancel.