53rd Putnam 1992

Show that the coefficient of x^k in the expansion of (1 + x + x² + x³)ⁿ is ∑_j=0^k  nCj   nC(k-2j).

Solution

1 + x + x² + x³ = (1 + x)(1 + x²). So the expression is (∑ nCi xⁱ) (∑ nCj x^2j). To get a term in x^k we must multiply a term x^2j by a term x^n-2j for some j. The result follows.

53rd Putnam 1992

© John Scholes
jscholes@kalva.demon.co.uk
7 Jan 2001