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Author: Tinku Tara

k-1-1-k-1-1-2k-1-2n-1-pi-2n-1-2-2n-2-2n-E-2m-

Question Number 222599 by MrGaster last updated on 01/Jul/25 $$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }=\frac{\pi^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{2}{n}\right)!}\mid{E}_{\mathrm{2}{m}} \mid \\ $$ Commented by MathematicalUser2357 last updated on…

sin-2-x-cos-x-1-3sin-2-x-dx-I-can-t-solve-this-maybe-some-genius-can-WolframAlpha-gives-a-solution-but-I-need-the-path-

Question Number 222595 by Ghisom last updated on 01/Jul/25 $$\Omega=\int\:\frac{\mathrm{sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cos}\:{x}}}{\mathrm{1}+\mathrm{3sin}^{\mathrm{2}} \:{x}}{dx} \\ $$$$\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{this},\:\mathrm{maybe}\:\mathrm{some}\:\mathrm{genius}\:\mathrm{can}. \\ $$$$\mathrm{WolframAlpha}\:\mathrm{gives}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{but}\:\mathrm{I} \\ $$$$\mathrm{need}\:\mathrm{the}\:\mathrm{path}… \\ $$ Commented by MrGaster last updated…

Question-222585

Question Number 222585 by Rojarani last updated on 30/Jun/25 Commented by AntonCWX8 last updated on 01/Jul/25 $${Now}\:{I}'{ll}\:{give}\:{you}\:{guys}\:{another}\:{challenge}. \\ $$$${What}\:{if}\:{the}\:{numbers}\:{beside}\:{the}\:\sqrt{\:\:}\:{symbol}\:{represent}\:{the}\:{nth}\:{root}? \\ $$$${Say}\:\mathrm{5}\sqrt{\:\:\:}\:=\:\sqrt[{\mathrm{5}}]{\:\:\:\:} \\ $$ Commented by…

If-a-i-gt-0-b-i-gt-0-i-1-n-Prove-that-a-1-2-b-1-2-a-2-2-b-2-2-a-n-2-b-n-2-a-1-a-2-a-n-2-b-1-b-2-b-

Question Number 222582 by hardmath last updated on 30/Jun/25 $$\mathrm{If}:\:\:\:\mathrm{a}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{b}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{i}\:=\:\overline {\mathrm{1},…,\mathrm{n}} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} }\:+\:\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} }\:+…+\:\sqrt{\mathrm{a}_{\boldsymbol{\mathrm{n}}}…

solve-for-p-q-s-in-terms-of-c-qs-q-sp-2-s-qs-q-sp-p-0-q-c-p-1-2-sp-q-q-cp-p-1-2-q-c-3-I-have-to-find-non-zero-real-x-q-c-p-1-

Question Number 222583 by ajfour last updated on 30/Jun/25 $${solve}\:{for}\:{p},{q},{s}\:{in}\:{terms}\:{of}\:{c}. \\ $$$$\bullet\:\left(\frac{{qs}}{{q}−{sp}}\right)^{\mathrm{2}} −{s}\left(\frac{{qs}}{{q}−{sp}}\right)+{p}=\mathrm{0} \\ $$$$\bullet\:\left(\frac{{q}+{c}}{{p}+\mathrm{1}}\right)^{\mathrm{2}} ={sp}−{q} \\ $$$$\bullet\:\left({q}−{cp}\right)\left({p}+\mathrm{1}\right)^{\mathrm{2}} =\left({q}+{c}\right)^{\mathrm{3}} \\ $$$${I}\:{have}\:{to}\:{find}\:{non}\:{zero}\:{real}\:{x}=−\left(\frac{{q}+{c}}{{p}+\mathrm{1}}\right)\:. \\ $$ Terms of…