Bionomials (1 Viewer)

andybandy · Oct 2, 2013

Hi guys,
I really need help with binomials, im struggling with them heaps, not the which term is independent of x stuff, but the usually question 6 or 7 of hsc extension papers, i always struggle or basically get them wrong, could anyone help explain, or help link me stuff that will help?

iBibah · Oct 2, 2013

andybandy said:
Hi guys,
I really need help with binomials, im struggling with them heaps, not the which term is independent of x stuff, but the usually question 6 or 7 of hsc extension papers, i always struggle or basically get them wrong, could anyone help explain, or help link me stuff that will help?

Binomial identities i presume ?

andybandy · Oct 3, 2013

Yeah, the prove that is that and hence that..

Drongoski · Oct 3, 2013

Best for you to post specific questions. No point having a very open-ended call for help so close to our HSC.

HAX0R · Oct 3, 2013

braintic · Oct 3, 2013

"Bionomials"

Are these numbers with a life of their own?

Or perhaps environmentally friendly numbers?

andybandy · Oct 3, 2013

HSC ext1 2008 question 6 c

edit - c instead of b

Drongoski · Oct 3, 2013

Andy - 2008 Q6(b) is not Binomial.

6(c)

i)

$(1+x)^{p+q} = \sum ^{p+q}_{i=0} \binom {p+q}{i} x^i\\ \\ \therefore \frac {(1+x)^{p+q}}{x^q} = \sum ^{p+q} _{i=0} \binom {p+q}{i}x^{i-q}\\ \\ $ The term indep of x is the one corresponding to i = q$ \\ \\ $This term is $ \binom {p+q}{q}$

ii) In the alternative formulation of (1+x)^(p+q)/x^q as (1+x)^p(1+1/x)^q:

$(1+x)^p (1+ \frac {1}{x})^q = \sum ^p _{i=0} \binom {p}{i}x^i \times \sum ^q _{j=0} \binom {q}{j} x^{-j} = (p+1) \times (q+1) $product terms of which only (p+1) are independent of x, being those where i=j and whose sum is $ \\ \\ \binom {p}{0} \binom {q}{0} + \binom {p}{1} \binom {q}{1} + \binom {p}{2} \binom {q}{2} + \dots + \binom {p}{p} \binom {q}{p} \\ \\ $and this is simply equal to the $ \binom {p+q}{q} $derived in i) above. $$

Bionomials (1 Viewer)

andybandy

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iBibah

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andybandy

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Drongoski

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