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Question Number 70834 by Maclaurin Stickker last updated on 08/Oct/19

Answered by mind is power last updated on 08/Oct/19

$$\sum _{k=1}^{n-1}\frac{1}{{log}_x\left(2^k\right){log}_x(2^{k+1)}}=\underset{k=1}{\sum ^{n-1}}\frac{1}{k(k+1){log}_x^2\left(2\right)}$$$$=\frac{1}{{log}_x^2\left(2\right)}\sum _{k=1}^{n-1}\frac{k+1-k}{k(k+1)}$$$$$$$$=\frac{1}{{log}_x^2\left(2\right)}\sum _{k=1}^{n-1}(\frac{1}{k}-\frac{1}{k+1})$$$$=\frac{1}{{log}_x^2\left(2\right)}(1-\frac{1}{n})=(1-\frac{1}{n})\times \frac{1}{{\log }_x^2\left(2\right)}$$$$=$$

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