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Author: Tinku Tara

Is-there-a-way-to-evaluate-the-following-S-2-2-2-2-2-

Question Number 2275 by Filup last updated on 13/Nov/15 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{way}\:\mathrm{to}\:\mathrm{evaluate}\:\mathrm{the}\:\mathrm{following}: \\ $$$$ \\ $$$${S}=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}} \\ $$ Answered by Rasheed Soomro last updated on 13/Nov/15 $${S}=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}}…

n-1-cos-n-pi-n-4-1-48pi-4-1-12-1-1-pi-2-pi-4-90-

Question Number 133344 by Dwaipayan Shikari last updated on 21/Feb/21 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left(\frac{{n}}{\pi}\right)}{{n}^{\mathrm{4}} }=−\frac{\mathrm{1}}{\mathrm{48}\pi^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{1}−\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\right)+\frac{\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$ Terms of Service Privacy Policy Contact:…

Sum-to-n-terms-1-1-1-1-2-1-1-2-3-

Question Number 2273 by RasheedAhmad last updated on 12/Nov/15 $${Sum}\:{to}\:{n}\:{terms}. \\ $$$$\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}… \\ $$ Answered by Yozzi last updated on 12/Nov/15 $${For}\:{an}\:{A}.{P}\:{with}\:{first}\:{term}\:{a}\:{and} \\ $$$${common}\:{difference}\:{d},\:{its}\:{sum}\:{to}\:{n} \\…

Find-modulus-and-argumen-of-z-1-i-4-3-i-7-1-i-2-8-1-i-3-12-

Question Number 133346 by bramlexs22 last updated on 21/Feb/21 $$\:\mathrm{Find}\:\mathrm{modulus}\:\mathrm{and}\:\mathrm{argumen}\:\mathrm{of}\: \\ $$$$\:\mathrm{z}\:=\:\frac{\left(\mathrm{1}−{i}\right)^{\mathrm{4}} \left(\sqrt{\mathrm{3}}+{i}\right)^{\mathrm{7}} }{\left(\mathrm{1}+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{8}} \left(−\mathrm{1}−{i}\sqrt{\mathrm{3}}\right)^{\mathrm{12}} } \\ $$ Answered by mathmax by abdo last updated…

Question-133341

Question Number 133341 by mohammad17 last updated on 21/Feb/21 Answered by mathmax by abdo last updated on 21/Feb/21 $$\left.\mathrm{1}\right)\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} \:+\mathrm{4x}^{\mathrm{2}} −\mathrm{10}\:\:\:\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=\mathrm{3x}^{\mathrm{2}} \:+\mathrm{8x}\:>\mathrm{0}\:\mathrm{on}\left[\mathrm{1},\mathrm{2}\right]\:\Rightarrow\mathrm{f}\:\mathrm{is}\:\mathrm{increazing} \\ $$$$\mathrm{on}\:\left[\mathrm{1},\mathrm{2}\right]\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{5}−\mathrm{10}=−\mathrm{5}<\mathrm{0}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{2}\right)=\mathrm{8}+\mathrm{16}−\mathrm{10}=\mathrm{14}>\mathrm{0}\:\Rightarrow…