What are the constants in the binomial series expansion for (1+mx)^-n?

[sooyong94]

I have pretty much the same notes for you:
Therefore:
... $-400-2+1\neq -399$ check arithmetic.

You have to remember to use what you learned in previous posts when you do future ones.
These lessons all fit together.

i.e. the numerator in the last calculation can be written in a shorter way remember?
This leads to a simpler way to write the whole fraction part.

Once you've done that - you need to compare what you got with what you have to show way back in post #1. If you go reread that but you should see the pattern already.

-401... Ok.. I get it now. :P

 

[Simon Bridge]

Great - so now all you have to do is put it all together!
Remember what you are trying to show?

 

[sooyong94]

Great - so now all you have to do is put it all together!
Remember what you are trying to show?

$\frac{(2)(3)(4)...(401)}{400!} (4)^{400}$

$=\frac{(1)(2)(3)(4)...(401)}{400!} 4^{400}$
$=\frac{(401!)}{400!} 4^{400}$
$=\frac{(401)(400!)}{400!} 4^{400}$
$=401 (4)^{400}$

$a=401, k=400$

 

[haruspex]

$\frac{(2)(3)(4)...(401)}{400!} (4)^{400}$

$=\frac{(1)(2)(3)(4)...(401)}{400!} 4^{400}$
$=\frac{(401!)}{400!} 4^{400}$
$=\frac{(401)(400!)}{400!} 4^{400}$
$=401 (4)^{400}$

$a=401, k=400$

That's it.

 

[Simon Bridge]

Beautiful - pour yourself a drink!