Using snake oil to evaluate $\sum_{k=0}^n(3k+1)\text C_n^k$

Question

I want to use snake oil to evaluate $\sum\limits_{k=0}^n(3k+1)\text C_n^k$. I want to try the method despite of the simplicity of this sum.

However, I got stuck after exchanging the order of summation.

Let $f(n)=\sum\limits_{k=0}^n(3k+1)\text C_n^k$, and $F(x)=\sum\limits_mx^mf(m)$. So \[F(x)=\sum_mx^m\sum_{k=0}^m(3k+1)\text C_m^k=\sum_k\sum_{m\ge k}x^m(3k+1)\text C_m^k.\] I don't know how to write this as a closed form.

Apparently it's a generating function technique: https://users.math.msu.edu/users/hensh/courses/482/spring19/lectures/14-SnakeOil.pdf — Karl, Jun 28 '23 at 16:11
It's a method in generating functions. EG Search for "Herb Wilf's Snake Oil Method" — Calvin Lin, Jun 28 '23 at 16:11
Off the top of my head, maybe try evaluating $\frac{d}{dx}[x(1+x^3)^n] |_{x=1}$ two different ways? — Daniel Schepler, Jun 28 '23 at 16:12
https://en.m.wikipedia.org/wiki/Generating_function#Introducing_a_free_parameter_(snake_oil_method) — RobPratt, Jun 29 '23 at 02:14

score 2 · Accepted Answer · answered Jun 28 '23 at 16:57

The factor $(3k+1)$ isn't involved when you are summing over $m$, so it can be pulled out of the inner summation. Next, assuming $C_n^k$ means ${n\choose k}$, you can use the formula $$ \sum_{m=k}^{\infty} \binom{m}{k} x^m = \frac{x^k}{(1-x)^{k+1}} $$ which you can prove by induction.

After simplification you'll have sums of the form $\sum_{k=0}^\infty r^k$ and $\sum_{k=0}^\infty kr^k$.

Using snake oil to evaluate $\sum_{k=0}^n(3k+1)\text C_n^k$

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