I came across this problem and I couldn't solve it
$\sum_{r=1}^\infty\frac{6^r}{(3^r-2^r)(3^{r+1}-2^{r+1})}$
So when I saw the solution they wrote $6^r = 3^r(3^{r+1}-2^{r+1}) - 3^{r+1}(3^r-2^r)$ and then applying the V(r)-V(r-1) to further solve it. I can't understand how they wrote $6^r$, I mean how did they construct it in the first place, what was the idea to form it?
And if possible is there any other elegant solution to this problem?
Note that $$T_r=\frac{6^r}{(3^r-2^r)(3^{r+1}-2^{r+1})}=\frac{3^r}{3^r-2^r}-\frac{3^{r+1}}{3^{r+1}-2^{r+1}}=F_r-F_{r+1}$$ By telescopic summation we get $$T_r=F_r-F_{r+1}$$ $$T_1=F_1-F_2$$ $$T_2=F_2-F_3$$ $$............$$ $$T_{n-1}=F_{n-1}-F_n$$ $$T_n=F_n-F_{n+1}$$ $$\implies S_n=\sum_{r=1}^{n}T_r=F_1-F_{n+1}=3-\frac{3^{n+1}}{3^{n+1}-2^{n+1}}$$ $$\implies S_{\infty}=3-1=2$$