Menu Close

Category: Algebra

Question-155183

Question Number 155183 by mathdanisur last updated on 26/Sep/21 Answered by aleks041103 last updated on 26/Sep/21 $$\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }=\frac{{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }−{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }}{{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{k}} }}=\frac{{cos}\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}}…

prove-that-log-2-2-2-n-1-n-n-1-2-2-n-log-2-n-1-n-n-1-3-2-n-n-1-n-n-1-4-2-n-23-8-6-2-2-3-1-18-log-6-2-m-A-

Question Number 155136 by amin96 last updated on 25/Sep/21 provethat$$\frac{\boldsymbol{\mathrm{log}}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\varphi}\left(\boldsymbol{{n}}\right)}{\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{2}^{\boldsymbol{{n}}} }+\boldsymbol{\mathrm{log}}\left(\mathrm{2}\right)\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\varphi}\left({n}\right)}{\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\mathrm{3}} \mathrm{2}^{\boldsymbol{{n}}} }+\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\varphi}\left({n}\right)}{\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\mathrm{4}} \mathrm{2}^{\boldsymbol{{n}}} }=…