(2^2-1)(2^2+1)(2^4+1)(2^8+1)
[spoiler]2^16-1[/spoiler]
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Cramya
i still do not understand this problem, can you please explain some more
thanks
i still do not understand this problem, can you please explain some more
thanks
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cramya
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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!
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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jnellaz
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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).
2^2-1 for example is equivalent to 2^2 minus 1 NOT 2^(2-1).
