<s>[INST] $2^{k} \mid {2k \choose 0} \cdot 3^{0} + {2k \choose 2} \cdot 3^{1} + \cdots + {2k \choose 2i} \cdot 3^{i} + \cdots + {2k \choose 2k} \cdot 3^{k}$ I have to prove this question:
$$2^{k} \,\left|\, {2k \choose 0} \cdot 3^{0} + {2k \choose 2} \cdot 3^{1} + \cdots + {2k \choose 2i} \cdot 3^{i} + \cdots + {2k \choose 2k} \cdot 3^{k}\right.$$
What i did was to get a closed form for this summation, and i haven't been able to get one. Any ideas, or is there any other method for proving this.
 [/INST] Let's talk about some general methods.  This is a special case of the following problem: given that you understand a sequence $a_n$, how can you understand the sequence $b_n = \sum_{k=0}^n {n \choose k} a_k$?  Generating function techniques will come to your rescue.  If you let $A(x) = \sum_{n \ge 0} a_n x^n$, then
$$B(x) = \sum_{n \ge 0} b_n x^n = \sum_{n \ge 0} \sum_{k=0}^{n} {n \choose k} a_k x^n.$$
At this point we employ one of the simplest but most effective tools in generatingfunctionology, namely we exchange the order of summation.  This gives
$$B(x) = \sum_{k \ge 0} \sum_{n \ge k} {n \choose k} a_k x^n.$$
Since we want an answer in terms of $A$, it seems natural to factor $a_k x^k$ out of the inner sum.  This gives
$$B(x) = \sum_{k \ge 0} a_k x^k \sum_{n \ge k} {n \choose k} x^{n-k}.$$
Now a generatingfunctionologist will immediately recognize the inner sum as $\sum_{m \ge 0} {m+k \choose k} x^m = \frac{1}{(1 - x)^{k+1}}$.  This gives
$$B(x) = \sum_{k \ge 0} a^k \frac{x^k}{(1 - x)^{k+1}} = \frac{1}{1 - x} A \left( \frac{x}{1 - x} \right).$$
Beautiful.  Now, in this problem we are dealing with the sequence $a_{2k} = 3^k, a_{2k+1} = 0$.  This has generating function $A(x) = \frac{1}{1 - 3x^2}$, hence
$$B(x) = \frac{1}{1 - x} \left( \frac{1}{1 - \frac{3x^2}{(1 - x)^2} } \right) = \frac{1 - x}{(1 - x)^2 - 3x^2} = \frac{1 - x}{1 - 2x - 2x^2}.$$
Now, what we want to prove is that the even coefficients of $B \left( \frac{x}{\sqrt{2}} \right)$ are integral.  To extract the even coefficients we appeal to another general tool in generatingfunctionology: if $B(x) = \sum_{n \ge 0} b_n x^n$ is a generating function you want to extract the even coefficients from, then these are given by
$$\sum_{n \ge 0} b_{2n} x^{2n} = \frac{B(x) + B(-x)}{2}.$$
This is a special case of the discrete Fourier transform.  Once you know this, it is completely mechanical to verify that
$$\frac{B \left( \frac{x}{\sqrt{2}} \right) + B \left( \frac{-x}{\sqrt{2}} \right)}{2} = \frac{1 - x^2}{1 - 4x^2 + x^4}$$ 
which has integer coefficients (since the constant term of the denominator is $1$).  This is a bit tedious, but my point here is that once one has picked up a few generating function techniques most of the work in solving problems like these is automatic.
</s>