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Author: Tinku Tara

mathematical-analysis-II-prove-that-0-1-1-1-x-ln-x-2-2x-1-1-x-x-2-n-1-1-n-2-2n-n-pi-2-18-

Question Number 137610 by mnjuly1970 last updated on 04/Apr/21 $$\:\:\:\:\:\:\:\:\:\:……..\:{mathematical}\:\:\:{analysis}\:\left({II}\right)…. \\ $$$$\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}}{ln}\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix}}=\frac{\pi^{\mathrm{2}} }{\mathrm{18}}.. \\ $$ Terms…

nice-calculus-prove-that-n-0-tan-1-1-F-n-tan-1-1-F-n-1-pi-2-4-F-n-is-fibonacci-sequence-

Question Number 137605 by mnjuly1970 last updated on 04/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:…..{nice}\:\:\:\:{calculus}… \\ $$$$\:\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{F}_{{n}} }\right).{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{F}_{{n}+\mathrm{1}} }\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:{F}_{{n}} \:{is}\:{fibonacci}\:{sequence}…. \\…

x-1-1-x-1-1-2-1-3-1-4-1-5-Is-the-following-true-1-x-1-2x-1-2x-1-2x-1-2x-2x-2x-1-2x-2-1-2x-2x-2x-1-4x-2-2x-2x-1-1-2x-2x-1-

Question Number 6535 by Temp last updated on 01/Jul/16 $$\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+… \\ $$$$\boldsymbol{\mathrm{Is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{true}}? \\ $$$$\therefore\Sigma\frac{\mathrm{1}}{{x}}=\Sigma\left(\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}{x}−\mathrm{1}+\mathrm{2}{x}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}{x}−\mathrm{2}+\mathrm{1}+\mathrm{2}{x}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{4}{x}−\mathrm{2}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\…

The-yearly-requirement-for-a-particular-product-is-16000-units-and-its-unit-price-is-100-00-if-the-holding-cost-is-10-of-the-unit-price-while-its-ordering-cost-is-50-00-calculate-the-amount-of-or

Question Number 6524 by Tawakalitu. last updated on 30/Jun/16 $${The}\:{yearly}\:{requirement}\:{for}\:{a}\:{particular}\:{product}\:{is}\:\mathrm{16000}\:{units} \\ $$$${and}\:{its}\:{unit}\:{price}\:{is}\:\$\mathrm{100}.\mathrm{00},\:{if}\:{the}\:{holding}\:{cost}\:{is}\:\mathrm{10\%}\:{of}\:{the} \\ $$$${unit}\:{price}\:{while}\:{its}\:{ordering}\:{cost}\:{is}\:\$\mathrm{50}.\mathrm{00}\: \\ $$$${calculate}\:{the}\:{amount}\:{of}\:{order}\:{per}\:{annum}. \\ $$$$ \\ $$ Terms of Service Privacy Policy…