# Partial fractions and binomial question!

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#2

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#3

Have you done the general binomial expansion in core 4 (it's C4 on my exam board) yet?

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(Original post by

Have you done the general binomial expansion in core 4 (it's C4 on my exam board) yet?

**Davelittle**)Have you done the general binomial expansion in core 4 (it's C4 on my exam board) yet?

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#5

for each one of them substitute in the right number up to 4 terms and you should get the answer

P.s.

remember that this is for (1+x)^n therefore you'll have to acknowledge -x and the x^2 :P

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#6

(Original post by

Yes, but it's only to the power of -1 so it will just be 1/1-x

**ThePersian**)Yes, but it's only to the power of -1 so it will just be 1/1-x

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(Original post by

The power is just a value for n in this case -1

**Username1818**)The power is just a value for n in this case -1

So, for both I should go to the point x^4

and then what is part 3 asking?

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#8

(Original post by

Okay, thanks.

So, for both I should go to the point x^4

and then what is part 3 asking?

**ThePersian**)Okay, thanks.

So, for both I should go to the point x^4

and then what is part 3 asking?

hey hey... sorry I just noticed.. the formula it's missing x^2 and x^3 (I got the picture from the internet and didn't notice the mistake)

But it basically goes

(1+x)^n = 1 + nx + ((n(n-1))*x^2)/2! +((n(n-1)(n-2))*x^3)/3! +....

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#9

For the expansion of the second one what do people get for the coefficient of x^6?

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#10

(Original post by

For the expansion of the second one what do people get for the coefficient of x^6?

**Davelittle**)For the expansion of the second one what do people get for the coefficient of x^6?

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#11

(Original post by

Personally I got 0... unless you weren't asking me xD then sorry

**Username1818**)Personally I got 0... unless you weren't asking me xD then sorry

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#12

(Original post by

Could you explain how you got 0, I think I'm being really stupid but its (-1)(-2)(-3)(X^2)^3/3! how does that give 0

**Davelittle**)Could you explain how you got 0, I think I'm being really stupid but its (-1)(-2)(-3)(X^2)^3/3! how does that give 0

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#13

(Original post by

Honestly I just got the answer from http://www.wolframalpha.com/ -> best website ever Because I just wanted to check the proof for part c But I know what you mean... you do get -1

**Username1818**)Honestly I just got the answer from http://www.wolframalpha.com/ -> best website ever Because I just wanted to check the proof for part c But I know what you mean... you do get -1

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#14

(Original post by

I know on wolfram alpha i saw it was 0 as well but I'm just wondering how they actually got it :L

**Davelittle**)I know on wolfram alpha i saw it was 0 as well but I'm just wondering how they actually got it :L

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#15

**Davelittle**)

I know on wolfram alpha i saw it was 0 as well but I'm just wondering how they actually got it :L

(Original post by

I'll ask my teacher and let you know tomorrow :P

**Username1818**)I'll ask my teacher and let you know tomorrow :P

You do realize that O(x^6) means "terms of order x^6 and higher" and not zero (0) don't you?

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#16

(Original post by

Are you misunderstanding the output from Wolfram?

You do realize that O(x^6) means "terms of order x^6 and higher" and not zero (0) don't you?

**davros**)Are you misunderstanding the output from Wolfram?

You do realize that O(x^6) means "terms of order x^6 and higher" and not zero (0) don't you?

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#17

(Original post by

Not until now xD

**Username1818**)Not until now xD

It's called "big oh" notation because the big O indicates that terms of the power in brackets and higher have been omitted from the expansion

(Remember: O (oh) does not equal 0 (zero) )

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#18

(Original post by

No problem!

It's called "big oh" notation because the big O indicates that terms of the power in brackets and higher have been omitted from the expansion

(Remember: O (oh) does not equal 0 (zero) )

**davros**)No problem!

It's called "big oh" notation because the big O indicates that terms of the power in brackets and higher have been omitted from the expansion

(Remember: O (oh) does not equal 0 (zero) )

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#19

**davros**)

No problem!

It's called "big oh" notation because the big O indicates that terms of the power in brackets and higher have been omitted from the expansion

(Remember: O (oh) does not equal 0 (zero) )

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