Question

N=(2+1)(2^2+1)(2^4+1).... (2^32+1) and N=2^x. Then find x

Answer 1

Hi ! there is a mistake in your question..

It should be
N=(2+1)(2^2+1)(2^4+1).... (2^32+1) + 1
--------------------

Answer after correcting the question...

N = 2^x = 2^64-1+1 => 2^64
hence your answer will be 64

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In the second last step I wrote 2^38 by mistake ... it is 2³²

Answer 2

The answer is given below :

Given,

N = (2 + 1) (2^2 + 1) (2^4 + 1) ... (2^32 + 1) + 1

= 1 × (2 + 1) (2^2 + 1) (2^4 + 1) ... (2^32 + 1) + 1

= {(2 - 1) (2 + 1)} (2^2 + 1) ... (2^32 + 1) + 1

= (2^2 - 1) (2^2 + 1) ... (2^32 + 1) + 1

Proceeding in this way, we will get

N = (2^64 - 1) + 1

= 2^64

Given, N = 2^x

So,

2^x = 2^64

Now, taking log with (base e) both side, we get

log (2^x) = log (2^64)

=> x log2 = 64 log2

=> x = 64

So, the value of x is 64.

Thank you for your question.