Binomial Theorem (1 Viewer)

goobi · May 29, 2012

$Consider\the\,identity\,(1-t)^n(1+t)^n\equiv (1-t^2)^n,\,where\,n\,is\,a\,positive\,integer,\,show\,that\,for\,integral\,values\,of\,r,\,$
$\sum_{k=0}^{2r}(-1)^k\binom{n}{k}\binom{n}{2r-k}=(-1)^r\binom{n}{r}\,\,\,\,provided\,\,\,\,0\leq r\leq \frac{1}{2}n$

Any help would be appreciated

fortune_cookie · May 29, 2012

EDIT: you compare coeffiecents of t^(2r). Can't be bothered, it's midnight.

Also, this is from the 2007 Catholic Trial HSC. If you have this question , you would already have the solution.

fortune_cookie · May 29, 2012

LHS = (nC0-nC1 t + nC2 t^2 -nC3 t^3 +nC4 t^4 - ...... +nCn (-t)^n) (nC0+nC1 t +nC2 t^2 + ....+ nCn t^n)

Coefficent of t^2 = nC0 nCn -nC1 nCn-1 +nC2 nCn-2 - .....

Carrotsticks · May 29, 2012

goobi · May 30, 2012

Carrotsticks said:

Thanks Carrotsticks

This probably sounds very stupid but I wonder how you multiply the two summations in the first line (as I haven't learnt it yet). Shouldn't the term $\binom{n}{k}$ be squared also?

Carrotsticks · May 30, 2012

goobi said:
Thanks Carrotsticks
This probably sounds very stupid but I wonder how you multiply the two summations in the first line (as I haven't learnt it yet). Shouldn't the term $\binom{n}{k}$ be squared also?

That does not sound stupid at all.

I didn't actually multiply the two summations explicitly, that would have been FAR too tedious and long. Which C(n,k) did you think should be squared? I ask this because I have a couple up there. Were you referring to the C(n,k) from the expansion of (1-t^2)^n?

All I did in the first line was write both of the series out in Sigma notation because I personally prefer Sigma notation over expanded notation (although occasionally expanded form has its advantages) and I wanted to express both series in a more 'compact' form (not to be confused with 'closed form').

All I did was pick out the necessary terms that yield t^2r.

ie: t^0 x t^2r = t^2r etc

And by doing so, we of course multiply the coefficients, which is exactly what I did after the ===> symbol.

goobi · May 30, 2012

Carrotsticks said:
That does not sound stupid at all.

I didn't actually multiply the two summations explicitly, that would have been FAR too tedious and long. Which C(n,k) did you think should be squared? I ask this because I have a couple up there. Were you referring to the C(n,k) from the expansion of (1-t^2)^n?

All I did in the first line was write both of the series out in Sigma notation because I personally prefer Sigma notation over expanded notation (although occasionally expanded form has its advantages) and I wanted to express both series in a more 'compact' form (not to be confused with 'closed form').

All I did was pick out the necessary terms that yield t^2r.

ie: t^0 x t^2r = t^2r etc

And by doing so, we of course multiply the coefficients, which is exactly what I did after the ===> symbol.

I was referring to the C(n,k) from the expansion of L.H.S.

Carrotsticks · May 30, 2012

No, it should not be squared.

I didn't actually multiply the two terms, hence it is not squared.

Even if I did multiply the two terms, it would be incorrect to say [C(n,k)]^2 because k is the 'bound variable'.

That's like me saying:

$\sum_{k=1}^{\infty} k \times \sum_{k=1}^{\infty} \frac{1}{k^3} = \sum_{k=1}^{\infty} k \times \frac{1}{k^3} = \sum_{k=1}^{\infty} \frac{1}{k^2}$

Which is most certainly incorrect.

goobi · May 30, 2012

Carrotsticks said:
No, it should not be squared.

I didn't actually multiply the two terms, hence it is not squared.

Even if I did multiply the two terms, it would be incorrect to say [C(n,k)]^2 because k is the 'bound variable'.

That's like me saying:

$\sum_{k=1}^{\infty} k \times \sum_{k=1}^{\infty} \frac{1}{k^3} = \sum_{k=1}^{\infty} k \times \frac{1}{k^3} = \sum_{k=1}^{\infty} \frac{1}{k^2}$

Which is most certainly incorrect.

So how did you get t^2k in the first line then if you "didn't actually multiply the two terms"? That's what confuses me

Carrotsticks · May 30, 2012

goobi said:
So how did you get t^2k in the first line then if you "didn't actually multiply the two terms"? That's what confuses me

I just used the definition of the Binomial Expansion.

$\\ (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \\\\ $Replace $ x $ with $ (-t^2): \\\\ (1-t^2)^n = \sum_{k=0}^{n} \binom{n}{k} (-t^2)^k = \sum_{k=0}^{n} \binom{n}{k} (-1)^k t^{2k}$

Fus Ro Dah · May 30, 2012

fortune_cookie said:
EDIT: you compare coeffiecents of t^(2r). Can't be bothered, it's midnight.

Also, this is from the 2007 Catholic Trial HSC. If you have this question , you would already have the solution.

Not necessarily. What if OP was given the question from his/her teacher in a revision booklet or something as such?

Bored_of_HSC · May 31, 2012

goobi said:
Thanks Carrotsticks
This probably sounds very stupid but I wonder how you multiply the two summations in the first line (as I haven't learnt it yet). Shouldn't the term $\binom{n}{k}$ be squared also?

Goobi pls

Trebla · Jun 1, 2012

Um...looking at the first line of Carrotstick's solutions, the expansion should look like this:

$(1 -t)^n(1+t)^n = \left(\displaystyle\sum_{i=0}^n \binom{n}{i}(-1)^i t^i \right)\left(\displaystyle\sum_{j=0}^n \binom{n}{j}t^j \right) = \displaystyle\sum_{i=0}^n \displaystyle\sum_{j=0}^n \binom{n}{i}\binom{n}{j}(-1)^i t^{i+j}$

then you examine every combination of i + j = 2r...

The reason for that you can multiply the terms together is because you are effectively multiplying two general terms together when you expand the thing manually and finding every variation of i + j where i = {0, 1, .... , n} and j = {0, 1, .... , n}. Another way to see this is to note that in general, the double sigma means that

$\displaystyle\sum_{i=1}^n\displaystyle\sum_{j=1}^n a_i a_j \\\\= \displaystyle\sum_{i=1}^n\left(\displaystyle\sum_{j=1}^n a_i a_j\right)$

Note that the $a_i$ term in the brackets do not vary with each j so it can be factored out of the sum so we get

$\displaystyle\sum_{i=1}^n \left(a_i\left[\displaystyle\sum_{j=1}^n a_j\right]\right)$

Now we can look at the sum outside as we vary i. Notice that the other summation term with j's in the square brackets does not change as we vary i, so we can factorise it outside the sum.

$\left[\displaystyle\sum_{j=1}^n a_j\right] \displaystyle\sum_{i=1}^n a_i$

Now it should be clear that this leads to

$\left(\displaystyle\sum_{i=1}^n a_i\right) \left(\displaystyle\sum_{j=1}^n a_j\right)$

Fus Ro Dah · Jun 1, 2012

I am confused Trebla. Is the above relevant to the question, or is it just to demonstrate to goobi what happens when you multiply two summations?

Trebla · Jun 1, 2012

Fus Ro Dah said:
I am confused Trebla. Is the above relevant to the question, or is it just to demonstrate to goobi what happens when you multiply two summations?

The latter...

iSplicer · Jun 1, 2012

goobi said:
$Consider\the\,identity\,(1-t)^n(1+t)^n\equiv (1-t^2)^n,\,where\,n\,is\,a\,positive\,integer,\,show\,that\,for\,integral\,values\,of\,r,\,$
$\sum_{k=0}^{2r}(-1)^k\binom{n}{k}\binom{n}{2r-k}=(-1)^r\binom{n}{r}\,\,\,\,provided\,\,\,\,0\leq r\leq \frac{1}{2}n$

Any help would be appreciated

goobi pls.

But yeah, nice work Carrotsticks =D

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