harder binomial (1 Viewer)

[OLDMAN]

Show that the coeff. of x^k in the expression,
(1+x+x^2+x^3)^n is SUM(j:0--->int(k/2))[nCj.nC(k-2j)] where int(k/2) is the highest integer <= to k/2.

 

[underthesun]

Just a bit curious, where did the i come from?

ok never mind, now that it's fixed

 

[wogboy]

Correct me if I'm wrong, but I think there's a potential problem with that question. namely the the term nC(k-2j), in that summation. What happens if k-2j is negative? (i.e. j>k/2) This means that the term nC(k-2j) is undefined (you cannot have one number "choose" a negative). I know that j << k, at all times, but it isn't true that j << k/2 at all times. (had to put << instead of < due to annoying vB code)

e.g. take for instance when n=1 and you want to find the coefficent k, of x (i.e. k=1).

k = sum (j:0-->1) 1Cj * 1C(1-2j)
= (1C0 * 1C1) + (1C1 * 1C(-1))
= 1 + 1*1C(-1)
= 1 + 1C(-1)

and 1C(-1) is undefined (according to my calculator anyway )

 

[OLDMAN]

wogboy is absolutely right.

The upper limit should be int(k/2)=the highest integer <=k/2.

I have edited the problem accordingly.

 

[freaking_out]

Originally posted by OLDMAN
wogboy is absolutely right.

The upper limit should be int(k/2)=the highest integer <=k/2.

I have edited the problem accordingly.

u've been a bit "off the ball" recently....

 

[Archman]

if you are still working on it, stop scrolling now

















(1+x+x^2+x^3)^n
=(1+x^2)^n*(1+x)^n
=(sum(p: 0 to n) (nCp*x^2p)) (sum(q: 0 to n) (nCq*x^q))
now for x^k, we can get it by choosing a term from the first sum and making up the number by choosing something corresponding from the second sum
say if we choose nCj*x^2j from the first sum, we hafta choose
nC(k-2j)*x^(k-2j) from the second sum.
since j can be anything from 0 to half of k (or (k-1)/2 if its odd)
the coefficient of x^k is
sum (j: 0 to int(k/2)) nCj*nC(k-2j)

 

[freaking_out]

Originally posted by Archman
if you are still working on it, stop scrolling now

don't worry, i gave up ages ago...

 

[OLDMAN]

__________________________________________________
freaking_out:
u've been a bit "off the ball" recently....
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Yeah, I know. Permutation/combination problem, then this, oldman might be showing his age.

btw, archman, well done.
What made this problem difficult was a missing step : (1+x+x^2+x^3)^n = (1+x^2)^n*(1+x)^n
otherwise it would sit comfortably in q7 ext1.

 

[freaking_out]

Originally posted by OLDMAN
...What made this problem difficult was a missing step : (1+x+x^2+x^3)^n = (1+x^2)^n*(1+x)^n
otherwise it would sit comfortably in q7 ext1.

is this transformation a famous one or somethin'??if not how do we meant to come up with that?

 

[Lazarus]

This is just a guess; working backwards...

(1 + x + x² + x³)ⁿ = (1 + x²)ⁿ * (1 + x)ⁿ

(1 + x + x² + x³)ⁿ = [(1 + x²)(1 + x)]ⁿ

1 + x + x² + x³ = (1 + x²)(1 + x)

RHS could have been found by determining the sum of the geometric series on the LHS? Alternatively, I suppose it could just be factorised directly... I'm not sure whether students would then realise to raise both sides to the power of n etc though.

 

[freaking_out]

Originally posted by Lazarus
This is just a guess; working backwards...

(1 + x + x² + x³)ⁿ = (1 + x²)ⁿ * (1 + x)ⁿ

(1 + x + x² + x³)ⁿ = [(1 + x²)(1 + x)]ⁿ

1 + x + x² + x³ = (1 + x²)(1 + x)

RHS could have been found by determining the sum of the geometric series on the LHS? Alternatively, I suppose it could just be factorised directly... I'm not sure whether students would then realise to raise both sides to the power of n etc though.

and to think this could b a 3u question.

 

[OLDMAN]

___________________________________________________
Quote freaking_out
is this transformation a famous one or somethin'??if not how do we meant to come up with that?
___________________________________________________

Don't worry. In the HSC there'll be a part
i) show that 1 + x + x^2 + x^3 = (1 + x^2)(1 + x)

and work forwards to what Laz showed backwards.

 

[ND]

Originally posted by Lazarus
I'm not sure whether students would then realise to raise both sides to the power of n etc though.

Come on, give us some credit.

 

[ND]

Originally posted by OLDMAN

Don't worry. In the HSC there'll be a part
i) show that 1 + x + x^2 + x^3 = (1 + x^2)(1 + x)

and work forwards to what Laz showed backwards.

But isn't that just a single step? I mean, isn't it like saying show x^2+x=x(x+1)? It's just one factorisation.

 

[OLDMAN]

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Quote ND :
But isn't that just a single step? I mean, isn't it like saying show x^2+x=x(x+1)? It's just one factorisation.
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True. But I've seen simpler. May not necessarily be above, perhaps an alternative : i) factorize 1+x+x^2+x^3

I do however want to assure students that examiners are generally fair and wouldn't pose the cryptic questions I have been posting here for practice.

 

[freaking_out]

Originally posted by OLDMAN
...I do however want to assure students that examiners are generally fair and wouldn't pose the cryptic questions I have been posting here for practice.

well us students think differently about the examiners.

 

[Lazarus]

Originally posted by ND
Come on, give us some credit.

Yes, alright, point conceded.

 

[underthesun]

Originally posted by freaking_out
well us students think differently about the examiners.

The teacher at my school gives this "horrible" image of examiners where they always want to take your marks away. The point being that our works should be as neat as possible, as clear as possible etc..