# Evaluating $(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$

I'm trying to evaluate the following expression$$(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$$

I'm not really used to these types of problems, so I first tried using logarithms but I'm not sure what to do from there. See:

Let $$P = (2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$$.

Then we have: $$\log_2{P} = \log_2((2^1 + 3^1)...(2^{64} + 3^{64}))$$ $$\log_2{P} = \log_2(2^1 + 3^1) + \log_2(2^2 + 3^2) + ... + \log_2(2^{64} + 3^{64})$$ $$P = 2^{2^1 + 3^1} + 2^{2^2 + 3^2} +... + 2^{2^{64} + 3^{64}}$$ $$P = 2^{2^1}2^{3^1} + 2^{2^2}2^{3^2} +... + 2^{2^{64}}2^{3^{64}}$$

Would factoring $$2^2$$ be of any use here?

• $P = 2^{2^1 + 3^1} + 2^{2^2 + 3^2} +... + 2^{2^{64} + 3^{64}}$ $\;-\;$ That doesn't follow (and doesn't help, anyway, since the best you can get there is the definition of $P$ you started with).
– dxiv
Sep 26, 2021 at 7:37
• It seems a bit silly to use '$\dots$' for omitting precisely one factor in the product. Sep 26, 2021 at 8:17
• Haha, sorry. The title to me felt like it was running out of space. Sep 26, 2021 at 8:34

Hint: Multiply by $$(3^1-2^1)=1$$
$$(3^1-2^1)(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\dots (2^{64} + 3^{64})\\ =(3^2-2^2)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\dots (2^{64} + 3^{64})\\=(3^4-2^4)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})\dots (2^{64} + 3^{64})$$
Sayan Dutta shows a clever way. Another way is to note that when you multiply the parentheses, each term will be of the form $$2^n 3^{127-n}$$: $$(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64}) = \prod_{k=0}^{6} (2^{2^k}+3^{2^k}) \\ = \sum_{n=0}^{127} 2^n 3^{127-n} = 3^{127} \sum_{n=0}^{127} (2/3)^n = 3^{127} \frac{1-(2/3)^{128}}{1-2/3} = 3^{128} \frac{1-(2/3)^{128}}{3-2} = 3^{128} - 2^{128}.$$