Quandaries and Queries


According to page 126 of Murtha & Willard's "Statistics and Calculus" (Prentice-Hall, 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.




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 xk in the expansion of (1+x)n is  C(n,k) = n!/k! (n-k)!. . (To make this true in all cases you have to agree to write 1 for 0! and for x0.)

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 xk in the expansion of (1+x)n is  C(n,k) = n!/k! (n-k)!. The task now is to show that in the expansion of (1+x)n+1 the coefficient of xk 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,

  1. 1 times the xk term in the expansion of (1+x)n
  2. x times the xk-1 term in the expansion of (1+x)n

According to the inductive assumption the coefficients are

  1. 1 C(n,k)
  2. 1 C(n,k-1)

Thus the coefficient of xk in the expansion of (1+x)n+1 is

C(n,k) + C(n,k-1)

= n!/k! (n-k)! + n!/(k-1)! (n-(k-1))!

= n!/k! (n-k)! + n!/(k-1)! (n-k+1)!

= n!(n-k+1)/k! (n-k)!(n-k+1) + n! k/k (k-1)! (n-k+1)!

= n!(n-k+1)/k! (n-k+1)! + n! k/k ! (n-k+1)!

= C(n+1,k)



Go to Math Central