Quandaries
and Queries 

According to page 126 of Murtha & Willard's "Statistics and Calculus" (PrenticeHall, 1973), Newton's binomial theorem can proved inductively. I suppose that was his method, which I would like to see. I am not knowledgeable enough to supply a rigorous proof, but the coefficients of the expansions do follow a satisfying pattern when (1+x)^{n+1} is compared to (1+x)^{n}. If calculus is required, I would like to know why. 

Hi, You certainly don't need calculus, just some facility in working with factorials. The theorem says that, if n is a positive integer and k is an integer with 0 ≤ k ≤ n, then the coefficient of x^{k} in the expansion of (1+x)^{n} is C(n,k) = ^{n!}/_{k! (nk)!. }. (To make this true in all cases you have to agree to write 1 for 0! and for x^{0}.) The induction is on n. Start the induction with n = 1. (1+x)^{1}= 1+x and the possible values of k are 0 and 1. You can check that C(1,0) and C(1,1) are both 1. For the inductive step assume that for a fixed value of n, the coefficient of x^{k} in the expansion of (1+x)^{n} is C(n,k) = ^{n!}/_{k! (nk)!}. The task now is to show that in the expansion of (1+x)^{n+1} the coefficient of x^{k} is C(n+1,k). (1+x)^{n+1} = (1+x)(1+x)^{n} and hence the term with x to the power k comes from two sources,
According to the inductive assumption the coefficients are
Thus the coefficient of x^{k} in the expansion of (1+x)^{n+1} is
Cheers, 

