Question Number 6351 by sanusihammed last updated on 24/Jun/16 $$\mathrm{1}\:+\:\mathrm{4}\:+\:\mathrm{9}\:+\:…….\:{n}^{\mathrm{2}} \:=\:\frac{{n}\left({n}\:+\:\mathrm{1}\right)\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)}{\mathrm{6}} \\ $$ Commented by prakash jain last updated on 24/Jun/16 $${proving}\:{for}\:{k}+\mathrm{1}\:{assuming}\:{result}\:{for}\:{k} \\ $$$$\frac{{k}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}{{k}}+\left({k}+\mathrm{1}\right)^{\mathrm{2}} \\…
Question Number 137420 by mnjuly1970 last updated on 02/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:……{mathematical}\:…\:…\:…\:{analysis}\left({II}\right)….. \\ $$$$\:\:\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\:\mathbb{R}} \left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{x}^{\mathrm{2}} \right)^{{n}} }{\left({n}!\right)^{\mathrm{2}} }\right){dx}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………………….. \\ $$ Commented…
Question Number 6350 by sanusihammed last updated on 24/Jun/16 $${Prove}\:{by}\:{matimatical}\:{induction} \\ $$$$\mathrm{1}\:+\:\mathrm{8}\:+\:\mathrm{27}\:+\:……\:+\:{n}^{\mathrm{3}} \:\:=\:\:\left[\frac{{n}\left({n}\:+\:\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$ Answered by prakash jain last updated on 24/Jun/16 $${n}=\mathrm{1} \\…
Question Number 6346 by sanusihammed last updated on 24/Jun/16 Commented by prakash jain last updated on 25/Jun/16 $$\mathrm{If}\:{a},{b},{c}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{there}\:\mathrm{are}\:\mathrm{no}\:{a},{b},{c}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{log}\:\frac{{a}}{{b}−{c}}=\mathrm{log}\:\frac{{b}}{{c}−{a}}=\mathrm{log}\:\frac{{c}}{{a}−{b}} \\ $$$${let}\:{there}\:\exists\:{a},{b},{c}\in\mathbb{R}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{condition} \\…
Question Number 137419 by mnjuly1970 last updated on 02/Apr/21 $$\:\:\:\:\:\:\:………{mathematical}\:\:\:\:….\:\:\:{analysis}…….. \\ $$$$\:\:\:\:\:\:\:{evaluate}…. \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{\mathrm{2}\pi{x}} −{e}^{\pi{x}} }{{x}\left(\mathrm{1}+{e}^{\mathrm{2}\pi{x}} \right)\left(\mathrm{1}+{e}^{\pi{x}} \right)}{dx}=\lambda\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right){dx}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\lambda\:=\:??? \\…
Question Number 6343 by sanusihammed last updated on 24/Jun/16 $$\int{x}^{\mathrm{3}} \:\sqrt{\mathrm{1}\:−\:{x}}\:\:{dx} \\ $$ Answered by nburiburu last updated on 24/Jun/16 $${by}\:{substitution}\:{t}=\sqrt{\mathrm{1}−{x}}\Rightarrow{x}=\mathrm{1}−{t}^{\mathrm{2}} \\ $$$${dx}=−\mathrm{2}{t}\:{dt} \\ $$$${I}=\int\left(\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 137412 by oustmuchiya@gmail.com last updated on 02/Apr/21 Answered by herbert last updated on 02/Apr/21 $${gradient}\:{of}\:{l}_{\mathrm{1}} \:=\:\frac{\mathrm{2}+\mathrm{4}}{\mathrm{5}+\mathrm{1}}\:=\frac{\mathrm{6}}{\mathrm{6}}=\mathrm{1} \\ $$$${but}\:{grad}\:{of}\:{l}_{\mathrm{1}} ×{l}_{\mathrm{2}} =−\mathrm{1} \\ $$$${grad}\:{of}\:{l}_{\mathrm{2}} =−\mathrm{1}…
Question Number 137415 by mr W last updated on 02/Apr/21 $${solve} \\ $$$${x}+\sqrt{{x}\left({x}+\mathrm{1}\right)}+\sqrt{{x}\left({x}+\mathrm{2}\right)}+\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}=\mathrm{2} \\ $$ Commented by MJS_new last updated on 03/Apr/21 $${x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{24}} \\…
Question Number 6340 by sanusihammed last updated on 24/Jun/16 Commented by nburiburu last updated on 24/Jun/16 $${seems}\:{it}\:{need}\:{more}\:{info}.\:{For}\:{the}\:{moment}\:{being}\:{could}\:{be}\:{that}\:{moving}\:{one}\:{unit}\:{from}\:{right}\:{to}\:{left}\:{adds}\:\mathrm{2}\:{units}\:{up}\:{and}\:{cost}\:\mathrm{1}\:{in}\:{the}\:{centre}. \\ $$$${However}\:{many}\:{other}\:{posibilities}\:{could}\:{explain}\:{the}\:{numbers}\:{in}\:{the}\:{diagram}. \\ $$ Commented by prakash jain…
Question Number 137410 by mohammad17 last updated on 02/Apr/21 $$\int_{\mathrm{1}} ^{\:\infty\:\:\:} \frac{{x}\mathrm{2}^{{x}} +\mathrm{7}}{\mathrm{3}^{{x}} +{lnx}+\mathrm{1}}\:{dx} \\ $$ Commented by mohammad17 last updated on 02/Apr/21 $$????\: \\…