Question Number 140506 by Dwaipayan Shikari last updated on 08/May/21 $${e}^{\left(\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{3}}+\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{4}}−\frac{\zeta\left(\mathrm{5}\right)}{\mathrm{5}}+…\right)} \:\:\:=? \\ $$ Answered by mnjuly1970 last updated on 08/May/21 Commented by mnjuly1970 last…
Question Number 9420 by tawakalitu last updated on 07/Dec/16 $$\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are}\:\mathrm{two}\:\mathrm{binary}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{are}\: \\ $$$$\mathrm{in}\:\mathrm{4}\:−\:\mathrm{bit}\:\mathrm{2}'\mathrm{s}\:\mathrm{complement}\:\mathrm{formate},\: \\ $$$$\mathrm{where}\:\mathrm{x}\:=\:\mathrm{00102}\:\mathrm{and}\:\mathrm{y}\:=\:\mathrm{11012}\::\:\mathrm{clearly}, \\ $$$$\mathrm{y}\:\mathrm{is}\:\mathrm{a}\:\mathrm{negative}\:\mathrm{number}.\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{result} \\ $$$$\mathrm{of}\:\mathrm{x}\:+\:\mathrm{y}\:\mathrm{in}\:\mathrm{decimal}\:\mathrm{formate}. \\ $$ Terms of Service Privacy Policy…
Question Number 9419 by tawakalitu last updated on 07/Dec/16 $$\mathrm{Develop}\:\mathrm{an}\:\mathrm{algorithm}\:\mathrm{and}\:\mathrm{draw}\:\mathrm{a}\:\mathrm{flowchat} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{1}\:−\:\mathrm{100} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 140487 by Dwaipayan Shikari last updated on 08/May/21 $$\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)\frac{{dx}}{{x}}={log}\left(\mathrm{2}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 139152 by TheSupreme last updated on 23/Apr/21 $${wich}\:{program}\:{to}\:{run}\:\:{genetic}\:{algorithm}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 73247 by Lontum Hans last updated on 09/Nov/19 Answered by Kunal12588 last updated on 09/Nov/19 $${a}_{\mathrm{8}} −{a}_{\mathrm{4}} =\mathrm{40}\:\:\:….\left(\mathrm{1}\right) \\ $$$${a}_{\mathrm{8}} =\frac{\mathrm{3}}{\mathrm{2}}{a}_{\mathrm{4}} \:\:\:\:\:\:\:…..\left(\mathrm{2}\right) \\…
Question Number 137575 by Engr_Jidda last updated on 04/Apr/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 137574 by Engr_Jidda last updated on 04/Apr/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 137347 by Dwaipayan Shikari last updated on 01/Apr/21 $$\frac{\mathrm{1}}{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}+…}}}}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\left(\pi+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}+…}}}=\frac{\mathrm{1}}{\pi}\:\: \\ $$$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{disprove}} \\ $$ Answered by MJS_new last updated…
Question Number 137059 by Dwaipayan Shikari last updated on 29/Mar/21 $$\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+..}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }+…}{\mathrm{4}^{\mathrm{4}} }+..=\frac{\mathrm{3}}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{8}} \\ $$ Terms of…