mindblanking. help for a simple binomial proofs Q (1 Viewer)

mreditor16 · Oct 4, 2014

okay here it is. I) and ii) were easy enough. just iii) is yeah annoying me.

thanks in advance.

faisalabdul16 · Oct 4, 2014

What paper?

mreditor16 · Oct 4, 2014

faisalabdul16 said:
What paper?

lol why? the full question is above.

anyway its north Sydney girls 2009. I looked at the solutions but I'm still confused....

faisalabdul16 · Oct 4, 2014

mreditor16 said:
lol why? the full question is above.

anyway its north Sydney girls 2009. I looked at the solutions but I'm still confused....

Every question you have posted or shown = new paper for me to do because i haven't seen it before

mreditor16 · Oct 4, 2014

faisalabdul16 said:
Every question you have posted or shown = new paper for me to do because i haven't seen it before

=)

but, knowing you, don't waste your time on the rest of the paper.

faisalabdul16 · Oct 4, 2014

mreditor16 said:
=)

but, knowing you, don't waste your time on the rest of the paper.

rest of the paper is straightforward?

Kurosaki · Oct 4, 2014

mreditor16 said:
lol why? the full question is above.

anyway its north Sydney girls 2009. I looked at the solutions but I'm still confused....

its not on my copy of the NSG MX2 trial for 09

Is this for MX1 instead?

Kurosaki · Oct 4, 2014

Nvm don't need the solutions when we can make our own

.

It is given that
$1-x+x^2-...+x^18=b_0+b_1w+b_2w^2+...+b_{18}w^{18}$
As w=x+1, then x=w-1

$1-x+x^2-x^3+...+x^{18}=1-(w-1)+(w-1)^2-...+(w-1)^{18}=b_0+b_1w+b_2w^2+...+b_{18}w^{18$

If we look only at the coefficients of w^2, we see that it equals: $\binom{2}{0}+\binom{3}{1}+\binom{4}{2}+...+\binom{18}{16}\\ $Using the fact that $ \binom{n}{k}=\binom{n}{n-k}$ this transforms into $ \\ \\ \binom{2}{2}+\binom{3}{2}+...+\binom{18}{2}= \binom{19}{3}$using the property of part (ii)$$

mreditor16 · Oct 4, 2014

Kurosaki said:
its not on my copy of the NSG MX2 trial for 09
Is this for MX1 instead?

oopsie, yes its in 3u :/

faisalabdul16 said:
rest of the paper is straightforward?

Yep! =)

Kurosaki said:
Nvm don't need the solutions when we can make our own .

It is given that
$1-x+x^2-...+x^18=b_0+b_1w+b_2w^2+...+b_{18}w^{18}$
As w=x+1, then x=w-1

$1-x+x^2-x^3+...+x^{18}=1-(w-1)+(w-1)^2-...+(w-1)^{18}=b_0+b_1w+b_2w^2+...+b_{18}w^{18$

If we look only at the coefficients of w^2, we see that it equals: $\binom{2}{0}+\binom{3}{1}+\binom{4}{2}+...+\binom{18}{16}\\ $Using the fact that $ \binom{n}{k}=\binom{n}{n-k}$ this transforms into $ \\ \\ \binom{2}{2}+\binom{3}{2}+...+\binom{18}{2}= \binom{19}{3}$using the property of part (ii)$$

thanks so much!

now I get it

mindblanking. help for a simple binomial proofs Q (1 Viewer)

mreditor16

Well-Known Member

faisalabdul16

Wot

mreditor16

Well-Known Member

faisalabdul16

Wot

mreditor16

Well-Known Member

faisalabdul16

Wot

Kurosaki

True Fail Kid

Kurosaki

True Fail Kid

mreditor16

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)