Show that the coefficient of xk in the expansion of (1 + x + x2 + x3)n is ∑j=0k nCj nC(k-2j).
1 + x + x2 + x3 = (1 + x)(1 + x2). So the expression is (∑ nCi xi) (∑ nCj x2j). To get a term in xk we must multiply a term x2j by a term xn-2j for some j. The result follows.
53rd Putnam 1992
© John Scholes
7 Jan 2001