How to Calculate 3^2048: Step-by-Step Guide

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To calculate 3^2048, the discussion emphasizes proving the equation (2+3)(2^2+3^2)...(2^2048 + 3^2048) + 2^4096 - 3^4096 = 0. Participants suggest using the difference of squares formula to simplify the expression, starting with a = 2^2048 and b = 3^2048. The method involves iterating through pairs of powers, ultimately leading to the conclusion that the signs must be carefully considered. Clarification is provided regarding the signs in the final expression, confirming the correct formulation. The discussion concludes with a consensus on the approach to arrive at the solution.
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the answer is 3^2048. How do I get there?
 
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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}## ...
 
i don't know, what can i do :S?
 
and the answer is not given, it's multiple choice
 
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.
 
Last edited:
but i got a plus sign not a minus sign...
 
tsuwal said:
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.
 
now i get it. thanks!
 
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