jmc

We can attempt to deconstruct the summand by applying supposing that it breaks down like a partial fraction: $$\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}=\frac{A}{3^k-2^k}+\frac{B}{3^{k+1}-2^{k+1}}.$$Then $$6^k=A(3^{k+1}-2^{k+1})+B(3^k-2^k),$$which expands as $$6^k=(3A+B)3^k-(2A+B)2^k.$$It makes sense to make both $(3A+B)3^k$ and $(2A+B)2^k$ multiples of $6^k$ that differ by $6^k.$ To this end, set $(3A+B)3^k=(n+1)6^k$ and $(2A+B)2^k=n{6}^k.$ Then $3A+B=(n+1)2^k$ and $2A+B=n{3}^k$. Subtracting these equations, we get $A=(n+1)2^k-n{3}^k.$ It follows that $B=3n{3}^k-2(n+1)2^k,$ which gives us $$\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}=\frac{(n+1)2^k-n{3}^k}{3^k-2^k}+\frac{3n{3}^k-2(n+1)2^k}{3^{k+1}-2^{k+1}}.$$We can try setting $n$ to different values, to see what we get. If we set $n=0,$ then we get $$\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}=\frac{2^k}{3^k-2^k}-\frac{2^{k+1}}{3^{k+1}-2^{k+1}},$$which makes the sum telescope.

Just to make sure the sum converges, we compute the $n$th partial sum: $$\begin{array}{rl}\sum _{k=1}^n\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}& =\sum _{k=1}^n\left(\frac{2^k}{3^k-2^k}-\frac{2^{k+1}}{3^{k+1}-2^{k+1}}\right)\\ & =2-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}\\ & =2-\frac{1}{(\frac{3}{2}{)}^{n+1}-1}.\end{array}$$As $n$ becomes very large, ${\left(\frac{3}{2}\right)}^{n+1}$ also becomes very large. Thus, the infinite sum is $\boxed{2}.$