Prove that for m = 5 (mod 6), mC2 - mC5 + mC8 - mC11 + ... - mC(m-6) + mC(m-3) ≠ 0.
Solution
Answer: (-3)(m-1)/2 + 1.
Let m = 6n+5. We have x(1 - x)m = mC0 x - mC1 x2 + mC3 x3 - ... - mCm xm+1 (*). Let ω be a complex cube root of 1. Substitute x = 1, ω and ω2 in (*) and add. The terms in x3r+1 and x3r+2 give zero since 1 + ω + ω2 = 0. So we get mC2 - mC5 + ... - mCm = ( ω(1 - ω)m + ω2(1 - ω2)m )/3.
Now (1 - ω2)m = (1 - ω)m(1 + ω)m = (1 - ω)m(-ω2)m = - ω(1 - ω)m (since m = 5 mod 6). So mC2 - mC5 + ... - mCm = ( ω(1 - ω)m - (1 - ω)m)/3 = - 1/3 (1 - ω)m+1. But (1 - ω)2 = 1 - 2ω + ω2 = - 3ω. So - 1/3 (1 - ω)m+1 = (-3)(m-1)/2.
The series in the question is missing the final term - mCm, so it sums to 1 + (-3)(m-1)/2, which is clearly never zero for m ≥ 5.
© John Scholes
jscholes@kalva.demon.co.uk
16 Jan 2001