Show that we cannot have 4 binomial coefficients nCm, nC(m+1), nC(m+2), nC(m+3) with n, m > 0 (and m + 3 ≤ n) in arithmetic progression.
Solution
Note that we can have three, for example: 7C1 = 7, 7C2 = 21, 7C3 = 35.
a, b, c are in arithmetic progression iff 2b = a + c. We have nCm = (m+1)/(n - m) nC(m+1), so 2 nCm+1 = nCm + nC(m+2) iff 2 = (n - m - 1)/(m + 2) + (m + 1)/(n - m). Simplifying, this becomes: m2 - (n - 2)m + (n2 - 5n + 2)/4 = 0. For any given n, this is a quadratic in m, so it has at most two solutions. If we have 4 consecutive coefficients in arithmetic progression, then the two solutions must differ by 1. If they are m1 and m2, then (m1 - m2) = 1 implies (m1 - m2)2 = 1 and hence (m1 + m2)2 - 4m1m2 = 1. So (n - 2)2 - (n2 - 5n + 2) = 1, and hence n = -1. Hence for n > 0 we cannot have four coefficients in arithmetic progression.
Alternatively, because nCm = nC(n-m), m and m+1 must be centrally placed (in other words n = 2m+1). Otherwise we would have four distinct solutions to the quadratic, which is impossible. But (2m+1)C(m-1), (2m+1)Cm, (2m+1)C(m+1), (2m+1)C(m+2) cannot be a solution because both (2m+1)C(m-1) and (2m+1)C(m+2) are less than (2m+1)Cm.
© John Scholes
jscholes@kalva.demon.co.uk
27 Jan 2001