(In reply to re: Trivial solution?
by Ady TZIDON)
So if we have (a + b + c)^n, then there will be all of the following terms in the expansion:
Basically, every combination of a, b, and c such that their exponents add up to n. There are (n+2) choose 2 such combinations, since we have to order the n powers and 2 partitions between the three components of the polynomial.
So if a polynomial of m terms is raised to the nth power, the number of terms in the expansion will be (n+m-1) choose (m-1).
So we have the number of terms in the expansion of P1, N(P1) = (n+m-1) choose (n-1) = (n+m-1)! / m!(n-1)!
and N(P2) = (n+m-1) choose (m-1) = (n+m-1)! / n!(m-1)!
This simplifies nicely to N(P1) = N(P2) * n / m
So if 2 < n < m, then n/m < 1 and thus N(P1) < N(P2) for all n,m (which agrees with my "trivial" observation).
Edited on July 16, 2014, 3:46 pm
Posted by tomarken
on 2014-07-16 14:05:56