Question Number 121659 by Dwaipayan Shikari last updated on 10/Nov/20 $$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{4}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{6}} {x}^{\mathrm{2}} \right)…..} \\ $$$$ \\ $$ Commented…
Question Number 56124 by Tawa1 last updated on 10/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19 $$\left.\mathrm{3}\right){cos}\theta=\frac{{x}_{\mathrm{1}} {x}_{\mathrm{2}} +{y}_{\mathrm{1}} {y}_{\mathrm{2}} }{\:\sqrt{{x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} }\:×\sqrt{{x}_{\mathrm{2}}…
Question Number 56120 by Tawa1 last updated on 10/Mar/19 Answered by 121194 last updated on 10/Mar/19 $$\mathrm{without}\:\mathrm{loss}\:\mathrm{of}\:\mathrm{generality},\:\mathrm{lets}\:\mathrm{takes}\:\mathrm{A}\:\mathrm{paralel}\:\mathrm{to} \\ $$$$\mathrm{axis}\:{x}\:\mathrm{and}\:\mathrm{B}\:\mathrm{dephased}\:\theta\:\mathrm{anti}−\mathrm{clockwise} \\ $$$${A}_{{x}} ={V} \\ $$$${A}_{{y}} =\mathrm{0}…
Question Number 121655 by TITA last updated on 10/Nov/20 Commented by Dwaipayan Shikari last updated on 10/Nov/20 $$\mathrm{1}=\mathrm{1}^{\mathrm{2}} \\ $$$$\mathrm{1}+\mathrm{3}=\mathrm{2}^{\mathrm{2}} \\ $$$$\mathrm{1}+\mathrm{3}+\mathrm{5}=\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}=\mathrm{4}^{\mathrm{2}} \\…
Question Number 121653 by Dwaipayan Shikari last updated on 10/Nov/20 $$\frac{{log}\mathrm{1}}{\:\sqrt{\mathrm{1}}}−\frac{{log}\mathrm{3}}{\:\sqrt{\mathrm{3}}}+\frac{{log}\mathrm{5}}{\:\sqrt{\mathrm{5}}}−\frac{{log}\mathrm{7}}{\:\sqrt{\mathrm{7}}}+…. \\ $$ Commented by TANMAY PANACEA last updated on 10/Nov/20 $${excellent}\:{question}\:{but}\:{still}\:{thinking}\:{how}\:{to}\:{start} \\ $$ Commented…
Question Number 187155 by TUN last updated on 14/Feb/23 $${f}\left({x}\right)={x}.{e}^{\mathrm{2}{x}} \\ $$$$=>{f}^{\left({n}\right)} \left({x}\right)= \\ $$$$ \\ $$ Commented by 0670322918 last updated on 14/Feb/23 Commented…
Question Number 121454 by Dwaipayan Shikari last updated on 08/Nov/20 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\left(\mathrm{1}.\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{2}.\mathrm{3}\right)^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\left(\mathrm{1}.\mathrm{2}.\mathrm{3}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{2}.\mathrm{3}.\mathrm{4}\right)^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\left(\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}\right)^{\mathrm{2}} }+….. \\ $$$$ \\ $$$${Or} \\ $$$$\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 186962 by mnjuly1970 last updated on 12/Feb/23 Answered by MJS_new last updated on 12/Feb/23 $$\sqrt{{a}}={p}\wedge\sqrt{{b}}={q}\:\Rightarrow\:{p},\:{q}\:>\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{{p}^{\mathrm{2}} }−{q}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{{q}^{\mathrm{2}} }−{p}−\mathrm{1}\right)\geqslant\frac{\mathrm{25}}{\mathrm{4}} \\ $$$${q}=\mathrm{1}−{p}\:\Rightarrow\:\mathrm{0}<{p}<\mathrm{1} \\ $$$$\left(\frac{\mathrm{1}}{{p}^{\mathrm{2}}…
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Question Number 55859 by Easyman32 last updated on 05/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19 $${S}_{{n}} =\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right] \\ $$$${S}_{{n}−\mathrm{1}} =\frac{{n}−\mathrm{1}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{2}\right){d}\right] \\ $$$${S}_{{n}−\mathrm{2}} =\frac{{n}−\mathrm{2}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{3}\right){d}\right] \\…