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by cramya » Fri Jan 02, 2009 2:43 pm
Keep applying (a+b) (a-b) = (a^2-b^2) till u get to the answer

(2^2-1)(2^2+1)(2^4+1)(2^8+1)

(2^2-1)(2^2+1) = 2^4-1

(2^4-1) (2^4+1) = 2^8-1


(2^8-1) (2^8+1) = 2^16-1

Hope this helps!
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explain some more

by [email protected] » Sun Jan 04, 2009 8:37 pm
Cramya

i still do not understand this problem, can you please explain some more

thanks
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by cramya » Sun Jan 04, 2009 8:42 pm
Keep applying (a+b) (a-b) = (a^2-b^2) till u get to the answer

First time a = 2^2 b =1

(a+b) (a-b) = (2^2-1)(2^2+1) = (2^4 -1)

{2^2 squared = (2^2)^2
Use exponent formula (x^y)^z = x^ (y*z) here x=2 y=2 z=2}


Second time a= 2^4 b=1

(a+b) (a-b) = (2^4-1)(2^4+1) = (2^8 -1)


Third time around a= 2^8 b=1

(2^8-1) (2^8+1) = 2^16-1


Hope this helps!
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by jnellaz » Tue Jan 06, 2009 9:27 am
Sophia, maybe this tidbit will help clarify Cramya's explanation because I read the question incorrectly when I first looked at it:

2^2-1 for example is equivalent to 2^2 minus 1 NOT 2^(2-1).
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