Find the $n^{th}$ partial sum of a telescoping series $$
a_k = \frac{6^k}{(3^{k+1}+2^{k+1})(3^k + 2^k)}
$$
Been stuck on this for a while, I started by looking at values of $a_k$ for varying values of $k$ and looking for a pattern, but haven't come to anything useful yet. The next step, I think, is to try and find a pattern in $\sum a_k$ for varying values of k, but I'm having trouble simplifying some of the crazy expressions that result from that. The second part of the question asks what is the sum to infinity, but I think that once I find the $k^{th}$ partial sum, I can find the limit as $k\rightarrow\infty$.
But yeah, and help with this problem would be appreciated.
 A: Hint: 
$$a_k=\frac{2^k}{3^k+2^k}-\frac{2^{k+1}}{3^{k+1}+2^{k+1}}.$$
A: This is just a supplement on how to obtain hint given by Qing Zhang. 
Consider the following 
$$\frac{6^k}{(3^{k+1}+2^{k+1})(3^k+2^k)} = \frac{A \cdot 3^k + B \cdot 2^k}{3^{k+1} +2^{k+1}} + \frac{C\cdot 3^k + D\cdot 2^k}{3^k+2^k}$$
where $A,B,C,D$ are constants to be found. 
By making the same denominator on right side fractions and comparing numerators, we have
$$6^k = (A\cdot 3^k + B\cdot 2^k)(3^k+2^k) + (C\cdot 3^k+D\cdot2^k)(3^{k+1}+2^{k+1})$$
$$=(A \cdot 3^{2k} + A \cdot 6^k + B\cdot 6^k + B\cdot 2^{2k}) + (3C\cdot 3^{2k} + 2C\cdot 6^{k} + 3D\cdot 6^k +2D\cdot 2^{2k}).$$
Therefore, by comparing coefficient of $6^k,2^{2k}$ and $3^{2k},$ we have 
$$1 = A + B + 2C+3D,$$
$$0 = B + 3C+2D,$$
$$0=A+3C.$$
By solving simultaneous equations, we have 
$$A=0,B=-2,C=0,D=1.$$
Therefore, 
$$\frac{6^k}{(3^{k+1}+2^{k+1})(3^k+2^k)}=\frac{2^k}{3^k+2^k}-\frac{2^{k+1}}{3^{k+1}+2^{k+1}}.$$
A: Thats really weird because Zhang solution is not the only one: 
$$a_k=\frac{3^{k+1}}{3^ {k+1}+2^{k+1}}-\frac{3^{k}}{3^{k}+2^{k}}.$$
$$\sum_{k=1}^{n}a_k=\frac {3^{n+1}}{3^{n+1}+2^{n+1}} -\frac 3 5$$
