# Partial fractions and binomial question!

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#1
Hello,

Any one understand 4 part 2? I don't get what it's asking!
0
7 years ago
#2
(Original post by ThePersian)
Hello,

Any one understand 4 part 2? I don't get what it's asking!
It's asking you to write down the first 4 terms in the expansions of two binomial expressions. The expansion is written in your textbook and formula book.
0
7 years ago
#3
Have you done the general binomial expansion in core 4 (it's C4 on my exam board) yet?
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#4
(Original post by Davelittle)
Have you done the general binomial expansion in core 4 (it's C4 on my exam board) yet?
Yes, but it's only to the power of -1 so it will just be 1/1-x
0
7 years ago
#5
(Original post by ThePersian)
Hello,

Any one understand 4 part 2? I don't get what it's asking!
One of the general formulas used in C4 for binomial expansion is

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
0
7 years ago
#6
(Original post by ThePersian)
Yes, but it's only to the power of -1 so it will just be 1/1-x
The power is just a value for n in this case -1
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#7
The power is just a value for n in this case -1
Okay, thanks.
So, for both I should go to the point x^4

and then what is part 3 asking?
0
7 years ago
#8
(Original post by 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! +....
0
7 years ago
#9
For the expansion of the second one what do people get for the coefficient of x^6?
0
7 years ago
#10
(Original post by Davelittle)
For the expansion of the second one what do people get for the coefficient of x^6?
Personally I got 0... unless you weren't asking me xD then sorry
0
7 years ago
#11
Personally I got 0... unless you weren't asking me xD then sorry
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
0
7 years ago
#12
(Original post by 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
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
0
7 years ago
#13
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
I know on wolfram alpha i saw it was 0 as well but I'm just wondering how they actually got it :L
0
7 years ago
#14
(Original post by Davelittle)
I know on wolfram alpha i saw it was 0 as well but I'm just wondering how they actually got it :L
I'll ask my teacher and let you know tomorrow :P
0
7 years ago
#15
(Original post by Davelittle)
I know on wolfram alpha i saw it was 0 as well but I'm just wondering how they actually got it :L

I'll ask my teacher and let you know tomorrow :P
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?
1
7 years ago
#16
(Original post by 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?
Not until now xD
0
7 years ago
#17
Not until now xD
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) )
0
7 years ago
#18
(Original post by 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) )
I'll remember that for the future ^.^
0
7 years ago
#19
(Original post by 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) )
Oh my gosh thank you so much, never realised this at all >.<
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