Werg22
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I was wondering, is there any way to prove the bominial theorem without the use of induction? Thanks.
Pascal's triangle. The proof that this is true for all n requires induction.ircdan said:sum(C(n,k)*x^k * y^(n-k), k = 0, 1, ..., n) = (x + y)^n
ircdan said:sum(C(n,k)*x^k * y^(n-k), k = 0, 1, ..., n) = (x + y)^n
You can give a combinatorial proof.
Pf.
Consider the n factors of the expansion
(x + y)^n = (x+y)(x+y) * * * (x + y)
Now in this expansion, the coefficient of x^ky^(n-k) is the number of different ways we can select k x's from the n available factors. The total number of such selections of size k from a collection of size n is C(n, k), and we're done.
Ok apparently my proof was way too vague. Here is an example to help explain what's going on. For notation below, when i say xx i mean x*x = x^2, so I'm multiplying.Werg22 said:Could you please explain "Now in this expansion, the coefficient of x^ky^(n-k) is the number of different ways we can select k x's from the n available factors."? How do you come to this conclusion without knowledge of the binominial theorem?
Ohhhhh I think I see what you mean.quasar987 said:Pascal's triangle. The proof that this is true for all n requires induction.![]()
ircdan said:Ohhhhh I think I see what you mean.
To prove C(n,k) is the number of r-combinations of a set of n elements you need to use the fact that P(n,r) is the number of r-permutations of a set with n elements.
To prove P(n,r) is the number of r-permutations of a set with n elements you need to use the generalized product rule.
But the catch is, to prove the generalized form of the product rule, you need to use inductionSo even though I technically didn't use induction, it seems C(n,k) relies on a result that must be proved by induction. Unless there is another way to prove the product rule without induction? I don't know if there is, if so, then everything my proof works and it works without any use of induction right?
You need to use induction to prove that result, it's called the product rule. We use it all the time, but the proof is by induction(at least the one I've seen). I don't know if there is another way to prove it. I think that's why quasar987 was saying, I'm not sure though heheWerg22 said:For the first the digit you have a choice of n number, the second digit a choice of n-1 and so on. This leads to the conclusion that there is n(n-1)...(n-k+1) possibilities.
You betWerg22 said:By the way, thanks for your explanation, that is an elegant proof.