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let N=(2+1) (2 2 +1) (2 3 +1) …....(2 32 +1) +1 then log 256 N =?

let N=(2+1) (22+1) (23+1) …....(232+1) +1 then          log256N =?

Grade:12th pass

2 Answers

jagdish singh singh
173 Points
4 years ago
\hspace{-0.7 cm}$ Let $P=(2+1)(2^2+1)(2^4+1)..........(2^{32}+1)\;,$ Then\\\\\\ $P = \frac{\overbrace{(2-1)(2+1)}_{(2^2-1)}(2^2+1)(2^4+1)..........(2^{32}+1)}{(2-1)} = \frac{2^{64}-1}{2-1} = 2^{64}-1$\\\\\\ So $N=P+1 = 2^{64}-1+1 = 2^{64}$\\\\ So $\log_{256}(N)= \log_{256}(2^{64}) = 64\cdot \log_{2^{8}}(2) = \frac{64}{{8}}\log_{2}(2) = 8.$
Kushagra Madhukar
askIITians Faculty 629 Points
10 months ago
Dear student,
Please find the answer to your question.
 
Multiply and divide the given terms of N by (2 – 1)
Now apply the identity (a – b)(a + b) = a2 – b2 to club two terms into one.
Keep doing this and you will be left with N = 264 = (28)8 = 2568
Hence, log256N = 8
 
Thanks and regards,
Kushagra

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