Evaluate $\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdots(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}$ Evaluate $$\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}.$$
I can't figure out where to start. I tried using logarithms but I couldn't get a pattern going. Any advice will be helpful, thanks in advanced.
 A: Hint:
$$\frac{(5+6)+5^2}{3^1}=12$$
$$\frac{(5+6)(5^2+6^2)+5^4}{3^2}=144=12^2$$
$$\frac{(5+6)(5^2+6^2)(5^4+6^4)+5^8}{3^4}=20736=12^4$$
There is a pattern there. See if you can prove that the pattern continues.

One way is to generalize how something like $(5+6)(5^2+6^2)+5^4$ squares to become equal to $(5+6)(5^2+6^2)(5^4+6^4)+5^8$. [Because then just square both sides of one of the above equations to get to the next equation.]
As in:
$$
\begin{align}
\left((5+6)(5^2+6^2)+5^4\right)^2&=(5+6)^2(5^2+6^2)^2+2(5+6)(5^2+6^2)5^4+5^8\\
&=(5+6)(5^2+6^2)\left[(5+6)(5^2+6^2)+2(5^4)\right]+5^8
\end{align}
$$
So now it is about generalizing that something like $(5+6)(5^2+6^2)+2(5^4)$ is the same as $(5^4+6^4)$. I think writing all the $6$'s as $5+1$ and using the binomial theorem might get you there.
A: Hint  -  telescope:
$$
\begin{align}
&\quad \frac{\big(\color{red}{(6-5)\cdot}(6+5)\big)\;(6^2+5^2)(6^4+5^4)\cdot\dots\cdot(6^{1024}+5^{1024})+5^{2048}}{\color{red}{(6-5)\cdot}3^{1024}} \\
&= \frac{\big((6^2-5^2)(6^2+5^2)\big)\;(6^4+5^4)\cdot\dots\cdot(6^{1024}+5^{1024})+5^{2048}}{3^{1024}} \\
&= \frac{\big((6^4-5^4)(6^4+5^4)\big)\;(6^8+5^8)\cdot\dots\cdot(6^{1024}+5^{1024})+5^{2048}}{3^{1024}} \\
 &= \;\ldots
\end{align}
$$
A: First we simplify the numerator of $$ \frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}.$$
$$(5+6)(5^2+6^2)=(6-5)(5+6)(5^2+6^2)=(6^4-5^4)$$
$$(5+6)(5^2+6^2)(5^4+6^4)=(6^4-5^4)((5^4+6^4)=(6^8-5^8)
\\.\\.\\ (5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})=(6^{2048}-5^{2048})$$
$$ \frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}=\frac {  6^{2048} }{ 3^{1024}} = 12^{1024} $$
