For a polynomial $(x_1+x_2+...+x_N)^{2k}$, I am trying to show that the number of fully even terms is $≤a_kN^k$, where $a_k$ only depends on $k$ and is constant for a constant $k$. When I say "fully even terms" I mean terms where only even exponents appear.
For every term, the exponents have to add up to 2k. So I broke it down into all possible exponent combinations. E.g., the combination where all exponents equal 2, and there are k terms. For each of these combinations, I count the number of terms. I determined that for a combination with $i$ elements, there are ${N \choose i}$ terms. So I determined the total number of terms is $\sum_{i=1}^t{N \choose i}$, where $t = min(N,k)$.
But I have no idea how to compare this to the exponential expression I want. Is what I've done so far correct? And how can I proceed?
The terms you want to count are the terms of the form $x_1^{2k_1}\cdots x_N^{2k_n}$ where $k_i \ge 0, i=1,...,N,$ and $\sum{k_i} = k.$ This is exactly the number of terms in $(x_1+x_2+ \cdots x_N)^k$ or $N^k.$ We get a one-to-one correspondence by dividing/multiplying all the exponents by $2.$
My comment about partitions was way off base. In partitions, the order does not matter, but here you want to distinguish between $x_1^4x_2^2$ and $x_1^2x_2^4,$ for example.