Evaluate $\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdots(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}$

Question

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.

Answer 1

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.

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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.

Answer 2

Hint  -  telescope:

$$ \begin{align} &\quad \frac{\big(\color{red}{(6-5)\cdot}(6+5)\big)\;(6^2+5^2)(6^4+5^4)\cdot\dots\cdot(6^{1024}+5^{1024})+5^{2048}}{\color{red}{(6-5)\cdot}3^{1024}} \\ &= \frac{\big((6^2-5^2)(6^2+5^2)\big)\;(6^4+5^4)\cdot\dots\cdot(6^{1024}+5^{1024})+5^{2048}}{3^{1024}} \\ &= \frac{\big((6^4-5^4)(6^4+5^4)\big)\;(6^8+5^8)\cdot\dots\cdot(6^{1024}+5^{1024})+5^{2048}}{3^{1024}} \\ &= \;\ldots \end{align} $$

Answer 3

First we simplify the numerator of $$ \frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}.$$

$$(5+6)(5^2+6^2)=(6-5)(5+6)(5^2+6^2)=(6^4-5^4)$$

$$(5+6)(5^2+6^2)(5^4+6^4)=(6^4-5^4)((5^4+6^4)=(6^8-5^8) \\.\\.\\ (5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})=(6^{2048}-5^{2048})$$ $$ \frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}=\frac { 6^{2048} }{ 3^{1024}} = 12^{1024} $$