Evaluate $\sum_{k=0}^{n} {3n \choose 3k}$ Evaluate $\sum_{k=0}^{n} {3n \choose 3k}$.
It's not hard to evaluate by putting roots of unity. 
Therefore, I would like to see some solutions using elementary mathematical stuffs but not roots of unity. 
And the answer is $\frac{1}{3}(2^{3n}+2(-1)^n)$
 A: We want to count how many subsets of $\{1,2,\ldots,3n\}$ have a number of elements that is a multiple of $3$. For any $m\in\{0,1,2\}$, let $S_m^n$ be the set of subsets of $\{1,2,\ldots,3n\}$ with $3k+m$ elements and let $C_m^n=\left|S_m^n\right|$. We have the following recurrence relations:
$$\begin{align*}C_0^{n+1} &=& 2C_0^{n}+3 C_2^{n}+3 C_1^n\\C_1^{n+1}&=&2C_1^n+3C_0^n+3C_2^n\\C_2^{n+1}&=&2C_2^{n}+3C_1^n+3C_0^n\end{align*}$$
given by joining a subset of $\{1,2,\ldots,3n\}$ with a (possibly empty) subset of $\{3n+1,3n+2,3n+3\}$. On the other hand $C_0^n+C_1^n+C_2^n = 8^n$, hence
$$ C_0^{n+1} = 3\cdot 8^n-C_0^n $$
and the clam $C_0^n = \frac{8^n+2(-1)^n}{3}$ is straightforward to prove by induction.
By setting $T(n)=C_0^n-\frac{1}{3}8^n$ the previous recurrence relation simply becomes
$$ T(n+1) = - T(n).$$
A: If you let $f(n)$ be the sum in question, then, knowing the answer, we see that $$f(n+1)=7f(n)+8f(n-1).$$
If you can prove this recursion directly from the definition of $f(n)$ then you are done.
Now, by Vandermond's identity, we have: $$\begin{align}\binom{3n+3}{3k}=&\binom{3n-3}{3k-6}+6\binom{3n-3}{3k-5}+15\binom{3n-3}{3k-4}+20\binom{3n-3}{3k-3}\\
&+15\binom{3n-3}{3k-2}+6\binom{3n-3}{3k-1}+\binom{3n-3}{3k}
\end{align}\tag{1}$$
And:
$$\binom{3n}{3k}=\binom{3n-3}{3k-3}+3\binom{3n-3}{3k-2}+3\binom{3n-3}{3k-1}+\binom{3n-3}{3k}$$
This means that:
$$f(n)=\sum_{k=0}^{n-1} 2\binom{3n-3}{3k} + 3\sum_{k=1}^{n-1}\left(\binom{3n-3}{3k-2}+\binom{3n-3}{3k-1}\right)$$
So $$7f(n)+8f(n-1)=\sum_{k=0}^{n-1}22\binom{3n-3}{3k} + 21\sum_{k=1}^{n-1}\left(\binom{3n-3}{3k-2}+\binom{3n-3}{3k-1}\right)$$
Combine the terms of $(1)$ to get the same expression. 
To complete the induction, show that the expression is true for $n=0,1$.
A: First, observe that
$$\sum\limits_{k=0}^n \binom{3n}{3k} = \sum\limits_{k=0}^n \binom{3n-3}{3k} + 3\binom{3n-3}{3k-1} + 3\binom{3n-3}{3k-2} + \binom{3n-3}{3k-3}$$
by repeated expansion using Pascal's recurrence. Also, due to the limits on $k$ and the fact that $\binom{n}{m} = 0$ whenever $m \leq 0$ or $m > n$, we can write this
$$\sum\limits_{k=0}^n \binom{3n}{3k} = 3\sum\limits_{k=0}^n \bigg{(} \binom{3n-3}{3k-1} + \binom{3n-3}{3k-2} \bigg{)} + 2\sum\limits_{k=0}^n \binom{3n-3}{3k-3}$$
Now, let $F_n = \sum\limits_{k=0}^n \binom{3n}{3k}$. We use this substitution and the fact that the given sum over two binomial coefficients spans all binomial coefficients except those with bottom part a multiple of 3 to obtain;
$$F_n = 3\bigg{(} 2^{3n-3} - F_{n-1} \bigg{)} + 2F_{n-1} \implies F_n + F_{n-1} = 3*2^{3n-3}$$
To solve a linear nonhomogeneous recurrence like this, we let $F_n = h_n + p_n$ where $h_n$ is the homogeneous solution and $p_n$ is a particular solution to the nonhomogeneous form.
Solving $h_n + h_{n-1} = 0$ yields $h_n = A*(-1)^n$. We assume $p_n$ is of the form $p_n = B*2^{3n}$ and we solve;
$$p_n + p_{n-1} = 3*2^{3n-3} \implies B2^{3n-3}(2^{3} + 1) = 3*2^{3n-3}$$
which simplifies to $B = \dfrac{1}{3}$, so that $p_n = \dfrac{1}{3} 2^{3n}$.
Using an initial value to solve for $A$ suffices to prove the same formula the author claims.
