C4 Binomial Expansion

Not sure how to do this question: ​(a) ​Given that find the first four terms in the series expansion​of in descending powers of x.​(b) ​By taking x = 200 use your series to find ,giving your answer to 7 decimal places. ​(c) ​Use your series to find to the same degree of accuracy. ​Anyone have any ideas about how to tackle this? Thank you

Umm what is the question? binomial?

Given that what? I think you're missing part of the question

Sorry, I've edited it now

Sorry guys the expansion didn't copy, I've edited it again

Wouldn't just be:

1 + (1/2)(-2/x) + (1/2)(-1/2)[(-2/x^2)/2] + (1/2)(-1/2)(-3/2)[(-2/x^3)/6)].... and so on? for a

then for b plug x in

Oh thanks, i wasn't sure whether you could do it like that since it's 2/x and the mod of x>2 (although idk how or if the latter affects it)

Oh thanks, i wasn't sure whether you could do it like that since it's 2/x and the mod of x>2 (although idk how or if the latter affects it)

Oh thanks, i wasn't sure whether you could do it like that since it's 2/x and the mod of x>2 (although idk how or if the latter affects it)

You have to put the condition of mod x otherwise you end up dividing by 0 - not good. It doesn't affect the expansion though

I'm not the brightest sooo

@Student403
@Dick Diamonds

I'm not the brightest sooo

@Student403
@Dick Diamonds

You have to put the condition of mod x otherwise you end up dividing by 0 - not good. It doesn't affect the expansion though

You have to put the condition of mod x otherwise you end up dividing by 0 - not good. It doesn't affect the expansion though

Ah I see, thank you

That is not the reason.

I appreciate the compliment but neither am I

@Zacken

Lol don't worry, I think you're right !

Oh thanks, i wasn't sure whether you could do it like that since it's 2/x and the mod of x>2 (although idk how or if the latter affects it)

Don't listen to that. It's wrong. The binomial expansion $(1+x)^n$ for non-natural powers is only valid for $|x| <1$. (that is the power series converges in this radius), now map $x \mapsto \left|\frac{2}{x}\right|$ and the same condition states that $|x|>2$.



Indeed, thank you for pointing that out. I really wish some people wouldn't blurt out what they say without knowing anything about it.