Binomial Theorem Proof (1 Viewer)

[underthebridge]

Hey, is it possible that we may be tested to proove the formula for co-efficient nCr from (1+x)^n.

There is a similar question that has been asked in the 2003 Western Region Trial:
Verify the Pascal's Triangle relationship for Binomial Expansions:
i.e. Show that nCr + nC(r+1) = (n+1)C(r+1)

This question was done by just using the combinations formula and simplifying. The proof I'm referring to however seemed more complicated. It involved the same simplifying at the end but I think you had to show that pascal's triangle was in fact that formula that they gave i nthe Western Region paper.

 

[KFunk]

Here's an example of where this can become useful. You could extend the logic of it to show that the coefficient of 1^{n- k}x^k = n!/k!(n-k)! = ⁿC_k.

Alternatively consider (1 + x)ⁿ = (1 + x)(1 + x)... (1 + x) --> n times. The number of ways you can 'create' an x^k term is the number of ways you can select k x's from n brackets. Consider each bracket like a selection where you can choose either an 'x' or a '1'. Since we have n places from which we can get an x term there are ⁿC_k ways to get x^k, hence the coefficient of x^k is ⁿC_k. This gives the expansion:

(1 + x)ⁿ = ∑ ⁿC_k.x^k (k=0 to n)

For that other pascal's triangle thing you can just use the definition of ⁿC_r and show it via algebra.

 

[underthebridge]

Yeh nice, I'll have to have a good look at your other post after the HSC. There was a question I saw a while ago on tri-nomial probability. With your thing we could just as well have n-nomial probability.

I think the proof I'm talking about was showing that the coefficients of the expansion could be represented by nCr, so it involved writing out the co-efficients in terms of n only. A pattern was developed by trying a few examples of n, like 3 and 4. I doubt it will be needed for tomoro, but i thought i'd better be safe.