# Binomial series expansion

## Homework Statement

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.

## Homework Equations

Binomial expansion

## 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... :(

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Simon Bridge
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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##.
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?

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

Simon Bridge
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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?

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

Simon Bridge
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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?

Where's the value of m?

That actually looks like (401!) ?

Simon Bridge
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Where's the value of m?
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 actually looks like (401!) ?
That's right! So to ratio comes to 401!/400! doesn't it? ... what is that equal to?

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

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

Simon Bridge
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I know that 401!/400!=(401x400!)/(400!)=401
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.

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.

I don't know about that though... :/

But how do I use that m=4?

Simon Bridge
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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.

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}##

haruspex
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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}##
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.

So that means... the coefficient of ##x^{400}## is
##\frac{(-2)(-3)(-4)...(-400-2+1)}{400!} (-4)^{400}## ?

haruspex
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So that means... the coefficient of ##x^{400}## is
##\frac{(-2)(-3)(-4)...(-400-2+1)}{400!} (-4)^{400}## ?
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.

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.

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?

haruspex
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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?
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.

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.

How does all terms in the expansion are positive?

haruspex
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How does all terms in the expansion are positive?
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.

Simon Bridge
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It may be easier to think of it like this ...
the product of N negative numbers is positive if N is even and negative if N is odd right?
i.e. if ##y=(-x_1)(-x_2)(-x_3)\cdots(-x_{N-1})(-x_N): x_i>0##
then y > 0 if N is even and y < 0 if N is odd.
Can you see why this is the case?

...similarly ##(-4)^N=4^N## if N is even and ##-4^N## if N is odd.

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Simon Bridge
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Back to this:
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}##
Good - so what would the coefficient of x^3 look like?

Follow the pattern to find the expression for the coefficient of x^400 - put in the values of n and m you found and remember posts 4 and 5.

It may be easier to think of it like this ...
the product of N negative numbers is positive if N is even and negative if N is odd right?
i.e. if ##y=(-x_1)(-x_2)(-x_3)\cdots(-x_{N-1})(-x_N): x_i>0##
then y > 0 if N is even and y < 0 if N is odd.
Can you see why this is the case?

...similarly ##(-4)^N=4^N## if N is even and ##-4^N## if N is odd.

I get it now... :D
Because when you multiply N negative number, and if N is an even number (in this case 400), the signs of each number cancel. :P

Back to this:

Good - so what would the coefficient of x^3 look like?

Follow the pattern to find the expression for the coefficient of x^400 - put in the values of n and m you found and remember posts 4 and 5.

So... that would look something like this...
the coefficient of ##x^{400}## is
##\frac{(-2)(-3)(-4)...(-400-2+1)}{400!} (-4)^{400}##

haruspex
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So... that would look something like this...
the coefficient of ##x^{400}## is
##\frac{(-2)(-3)(-4)...(-400-2+1)}{400!} (-4)^{400}##

Yes, but you should simplify the signs, as you've been shown.