number theory?

Question

prove by binomial theroem

(n choose 0) - 1/2(n choose 1) + 1/3(n choose 2) -............+
[(-1)^n]/n+1 (n choose n) = 1/(n+1)

Answer 1

Certainly, Ranto's answer is perfect, but he does not use the Binomial Theorem, as you were asked to do. Consider the following.

Summation (from j = 0 to j = n) [(nCj)*((-t)^j)*1^(n-j)] = (1-t)^n, by the Binomial Theorem. Now integrate both sides with respect to t, from 0 to 1. The antiderivatives are
summation (j=0 to j=n)[(nCj)*((-1)^j)*(t^(j+1))/(j+1) =
-[(1-t)^(n+1)]/(n+1). The evaluations of the antiderivatives from 0 to 1 give exactly the theorem you are trying to prove.

Answer 2

Prove it by induction. Assume that it is true for n -- then show it is true for (n+1). This isn't hard to do. Then show that it holds for n=1.