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Category: Arithmetic

prove-that-2-is-irrationl-number-

Question Number 45712 by malwaan last updated on 15/Oct/18 $$\mathrm{prove}\:\mathrm{that}\:: \\ $$$$\sqrt{\mathrm{2}}\:\mathrm{is}\:\mathrm{irrationl}\:\mathrm{number} \\ $$ Commented by maxmathsup by imad last updated on 15/Oct/18 $${let}\:{suppose}\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}}\:{with}\:{p}\:{and}\:{q}\:{integrsn}\:{natural}\:{and}\:\Delta\left({p},{q}\right)=\mathrm{1}\:\Rightarrow \\…

Question-111218

Question Number 111218 by Lekhraj last updated on 02/Sep/20 Answered by nimnim last updated on 02/Sep/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\mathrm{9}\:\:\:\:\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\mathrm{9}\:\:\:\:\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\mathrm{9}\:\:\:\:\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{4}\:\:\:\:\:\mathrm{9}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{5}\:\:\:\:\mathrm{9}\:\:\:\:\:\mathrm{7}…

prove-that1-3-2-3-3-3-n-3-n-2-n-1-2-4-for-every-natural-number-n-

Question Number 45585 by pieroo last updated on 14/Oct/18 $$\mathrm{prove}\:\mathrm{that1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +…+\mathrm{n}^{\mathrm{3}} =\frac{\mathrm{n}^{\mathrm{2}} \left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:\mathrm{for}\:\mathrm{every} \\ $$$$\mathrm{natural}\:\mathrm{number}\:\boldsymbol{\mathrm{n}} \\ $$ Commented by maxmathsup by imad…

Prove-that-p-n-a-1-a-2-a-n-n-n-a-1-a-2-a-n-n-N-

Question Number 45353 by pieroo last updated on 12/Oct/18 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{p}\left(\mathrm{n}\right)=\frac{\boldsymbol{\mathrm{a}}_{\mathrm{1}} +\boldsymbol{\mathrm{a}}_{\mathrm{2}} +…+\boldsymbol{\mathrm{a}}_{\mathrm{n}} }{\mathrm{n}}\:\geqslant\:^{\mathrm{n}} \sqrt{\boldsymbol{\mathrm{a}}_{\mathrm{1}} \boldsymbol{\mathrm{a}}_{\mathrm{2}} …\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} } \\ $$$$\forall\:\mathrm{n}\:\in\boldsymbol{\mathrm{N}} \\ $$ Commented by pieroo last…

1-2-2-3-3-4-4-5-

Question Number 176054 by BaliramKumar last updated on 11/Sep/22 $$\frac{\mathrm{1}}{\mathrm{2}!}\:+\:\frac{\mathrm{2}}{\mathrm{3}!}\:+\:\frac{\mathrm{3}}{\mathrm{4}!}\:+\:\frac{\mathrm{4}}{\mathrm{5}!}\:+\:………….\:\infty\:=\:? \\ $$ Answered by Ar Brandon last updated on 11/Sep/22 $${S}=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{n}−\mathrm{1}}{{n}!}=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}−\underset{{n}=\mathrm{2}}…