Finding a formula for the kth partial sum

Question

I need help finding the kth partial sum of the question below:

Consider the series $\sum_{n=1}^{\infty}a_n$ where $a_n=\ln\left(\frac{6n-1}{6n+5}\right)$. Find a formula, in closed form, for the $k^{th}$ partial sum $s_k$.

I think this is a telescoping series but I'm sorta lost on how to answer the question. Any help would be appreciated

Picture of Question

Answer 1

$$\begin{aligned}a_n \;=\;& \ln\left(\frac{6n-1}{6n+5}\right) \;=\; \ln(6n-1) \;-\; \ln(6n+5) \\[5mm] =\;& \ln(6n-1) \;-\; \ln(6(n+1)-1)\end{aligned}$$

Which means that $\sum a_n$ is a telescoping sum, giving us:

$$\sum^n_{k=1}a_k \;=\; \ln(5) \;-\; \ln(6(n+1)-1)$$