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# Really Weird Binomial watch

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1. I probably could do this...the really long way...but when this question says 'as a product of primes' - I have no idea!!

a: By considering the general term, find the coefficient of the term x^14 in the expansion of:

[x^3 + (3/x^2)]

b: Write down the expansion of (1 + x)^6. Hence by letting x= z + z^2 expand (1 + z + z^2) in ascending powers of z as far as the term in z^4

or just look at the attachement.

2. Work out what term in the expansion will give you x^14. Start from (x^3)^9 and (3/x^2)^9 as halfway, and work from there.

Then use nCr where n is 18 and r is the power you found above

Break your co-efficient down into its prime factors

b is pretty straightforward. Expand, substitute and stop when the powers hit z^4. Simplify
3. (Original post by Trangulor)
Work out what term in the expansion will give you x^14. Start from (x^3)^9 and (3/x^2)^9 as halfway, and work from there.

Then use nCr where n is 18 and r is the power you found above

Break your co-efficient down into its prime factors

b is pretty straightforward. Expand, substitute and stop when the powers hit z^4. Simplify
I really, really appreciate your help...but now I'm even more confused
4. Could you/anyone clarify
5. Ok. The standard binomial is:

(a+b)^n=a^n + na^(n-1) b + 0.5n(n-1)a^(n-2) b^2 + ... + b^n

Firstly, ignore the co-efficients and just concentrate on the powers. How can you get x^14 from (x^3)^p times (3/x^2)^q, where p+q=18?

Spoiler:
Show
By laws of indices, x^3p times x^-2q = x^14

3p-2q=14

2p + 2q = 36 (double p+q=18)

p=10, q=8

with these numbers, work out the x^14 term and its co-efficient, which you need to write as its prime factors
6. (Original post by Trangulor)
Ok. The standard binomial is:

(a+b)^n=a^n + na^(n-1) b + 0.5n(n-1)a^(n-2) b^2 + ... + b^n

Firstly, ignore the co-efficients and just concentrate on the powers. How can you get x^14 from (x^3)^p times (3/x^2)^q, where p+q=18?

Spoiler:
Show
By laws of indices, x^3p times x^-2q = x^14

3p-2q=14

2p + 2q = 36 (double p+q=18)

p=10, q=8

with these numbers, work out the x^14 term and its co-efficient, which you need to write as its prime factors
ok now your gunna think that im a complete idiot! ive never seen that formula - could you write it down in an attachment becuase i dont understand whats going on.

also i dont seem to understand what the answer is
7. Bah, edexcel website is down

The second one may be the more familiar, requires the nCr button

Edit: The formula is in the Edexcel formula booklet, so when the website is back up, google for it. Might want to give it a go now, as it may just be my internet playing up
Attached Images

8. (Original post by Trangulor)
Bah, edexcel website is down
Here are the formulae
Attached Images

9. Thanks Steve

NM, the formula is on Page 7, so you won't need to remember it

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Updated: October 12, 2006
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