Proof of binomial theorem by mathematical induction in Jones and Couchman 3U Math Bk2 (1 Viewer)

[Alej]

I'm reading through the proof of the binomial theorem by mathematical induction given in Jones and Couchman 3 Unit Mathematics Book 2.

In the course of the proof they have a fraction with k[(n-1)(n-2)...(n-k+1)]+[(n-1)(n-2)...(n-k)] in the numerator. The denominator is k!. In the next line of the proof the numerator is given as [(n-1)(n-2)...(n-k+1)][k+(n-k)] and the denominator is still k!.

I can't see how they arrived at this result. It seems to me that since (n-1)(n-2)...(n-k+1) is (n-1)(n-2)...(n-k)(n-k+1), the common factor here is (n-1)(n-2)...(n-k), so that k[(n-1)(n-2)...(n-k+1)]+[(n-1)(n-2)...(n-k)] = k(n-k+1)[(n-1)(n-2)...(n-k)]+[(n-1)(n-2)...(n-k)] = [(n-1)(n-2)...(n-k)][k(n-k+1)+1].

Where am I going wrong?

 

[InteGrand]

Re: Proof of binomial theorem by mathematical induction in Jones and Couchman 3U Math

I'm reading through the proof of the binomial theorem by mathematical induction given in Jones and Couchman 3 Unit Mathematics Book 2.

In the course of the proof they have a fraction with k[(n-1)(n-2)...(n-k+1)]+[(n-1)(n-2)...(n-k)] in the numerator. The denominator is k!. In the next line of the proof the numerator is given as [(n-1)(n-2)...(n-k+1)][k+(n-k)] and the denominator is still k!.

The numerators are the same, they just factored out (n-1)(n-2)...(n-k+1).

 

[Alej]

Re: Proof of binomial theorem by mathematical induction in Jones and Couchman 3U Math

Okay, I got it. I've been looking at it as (n-1)(n-2)...[n-(k+1)].