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Grade: 12th pass
        
let N=(2+1) (22+1) (23+1) …....(232+1) +1 then          log256N =?
3 years ago

Answers : (1)

jagdish singh singh
173 Points
							
\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 $\log_{256}(N)= \log_{256}(2^{64}) = 64\cdot \log_{2^{8}}(2) = \frac{64}{{8}}\log_{2}(2) = 8.$
3 years ago
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