I just saw a proof of the binomial theorem using the Taylor series. Is there some other proof of this series that doesn't involve f'(x)=nxⁿ -1, where f(x)=xⁿ , a result obtained using the binomial theorem?

Brad

You can prove this by induction assuming n is a positive integer, which I think you know about. If this is the theorem you're trying to prove, then just work out (x+a)(x+a)ⁿ , assuming the binomial theorem is true for n to expand (x+a)ⁿ . The actual proof is an exercise for the reader I think, however write another message if you're looking for a more general form of the binomial theorem (negative n or noninteger n).

Thanks, I shouldn't have overlooked that. But anyways, what is the more general form? (I didn't know that you could apply the theorem to non-integer values of n.)

Brad

(x+a)-n=

¥ å k=0

(-1)k n+k-1Ck xk a-n-k

And for non-integer n, you get for |x| < 1 that:

(x+1)ⁿ=1+n x+(n(n-1)/2)x²+(n(n-1)(n-2)/6)x³+¼