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Grade 12th passAlgebra

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

Pranav Dabhade

9 Years agoGrade 12th pass

2 Answers

jagdish singh singh

Approved Tutor Answer9 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

5 Years 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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