How do I calculate 3^2048?

[tsuwal]

the answer is 3^2048. How do I get there?

 

[AlephZero]

Since you are given the answer, use that information!

You have to prove that
$(2+3)(2^2+3^2)\cdots(2^{2048} + 3^{2048}) + 2^{4096} - 3^{4096} = 0$
Now, think what you can do with $2^{4096} - 3^{4096}$ ...

 

[tsuwal]

i don't know, what can i do :S?

 

[tsuwal]

and the answer is not given, it's multiple choice

 

[mathman]

a² - b² = (a-b)(a+b)

Start with a = 2²⁰⁴⁸ and b = 3²⁰⁴⁸
next repeat with a = 2¹⁰²⁴ and b = 3¹⁰²⁴
etc.
At the end you will have (2-3)(2+3). Just be careful with the sign.

 

[tsuwal]

but i got a plus sign not a minus sign...

 

[HallsofIvy]

but i got a plus sign not a minus sign...

No, its a minus sign:
$(2+3)(2^2+3^2)\cdots(2^{2048} + 3^{2048}) + 2^{4096} - 3^{4096} = 0$
AlephZero was referring to the last pair on the left.

 

[tsuwal]

now i get it. thanks!