Question Number 138789 by 676597498 last updated on 18/Apr/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 7723 by FilupSmith last updated on 12/Sep/16 $${p}_{{n}} ={n}\mathrm{th}\:\mathrm{prime}\:\:\left({p}_{\mathrm{1}} =\mathrm{2},\:\:{p}_{\mathrm{2}} =\mathrm{3},\:\:\:{p}_{\mathrm{3}} =\mathrm{5},\:…\right) \\ $$$$\mathrm{Do}\:\mathrm{the}\:\mathrm{following}\:\mathrm{sums}\:\mathrm{converge}?\:\mathrm{Prove}/\mathrm{disprove}. \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{{p}_{{n}} } \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{{p}_{{n}}…
Question Number 7656 by Tawakalitu. last updated on 07/Sep/16 $${If}\:\:{S}_{{n}} \:\left({sum}\:{of}\:{sequence}\right)\:=\:\mathrm{2}{n}^{\mathrm{2}} \:+\:{n} \\ $$$${Find}\:{the}\:{value}\:{of}\:{the}\:\mathrm{1002}\:{term} \\ $$ Commented by prakash jain last updated on 12/Sep/16 $${n}^{{th}}…
Question Number 138698 by mathlove last updated on 16/Apr/21 Answered by nimnim last updated on 16/Apr/21 $$\frac{{a}}{{b}}=\mathrm{1}.\mathrm{454545}…..=\frac{\mathrm{145}−\mathrm{1}}{\mathrm{99}}=\frac{\mathrm{144}}{\mathrm{99}}=\frac{\mathrm{16}}{\mathrm{11}} \\ $$$$\Rightarrow{a}=\mathrm{16},\:{b}=\mathrm{11} \\ $$$$\therefore\:{a}+{b}=\mathrm{16}+\mathrm{11}=\mathrm{27} \\ $$ Commented by…
Question Number 7609 by Tawakalitu. last updated on 06/Sep/16 $${Divide}\: \\ $$$${a}^{\mathrm{5}/\mathrm{2}} \:−\:\mathrm{5}{a}^{\mathrm{2}} {b}^{\mathrm{1}/\mathrm{3}} \:+\:\mathrm{10}{a}^{\mathrm{3}/\mathrm{2}} \mathrm{6}^{\mathrm{2}/\mathrm{3}} \:−\:\mathrm{10}{ab}\:+\:\mathrm{5}{a}^{\mathrm{1}/\mathrm{2}} {b}^{\mathrm{4}/\mathrm{3}} \:−\:{b}^{\mathrm{5}/\mathrm{3}} \\ $$$${by}\:\:{a}^{\mathrm{1}/\mathrm{2}} \:−\:{b}^{\mathrm{1}/\mathrm{3}} \\ $$ Answered…
Question Number 7590 by FilupSmith last updated on 05/Sep/16 $${S}\:=\:\underset{{n}=\mathrm{2}} {\overset{{k}} {\sum}}\:\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{{n}\left({n}−\mathrm{1}\right)} \\ $$$${S}=? \\ $$ Commented by Yozzia last updated on 05/Sep/16 $$\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{{n}\left({n}−\mathrm{1}\right)}\equiv\frac{{a}}{{n}}+\frac{{b}}{{n}−\mathrm{1}} \\…
Question Number 7545 by Tawakalitu. last updated on 03/Sep/16 $${y}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} \:=\:\mathrm{2}^{{x}} \:+\:\mathrm{2}^{{y}} \\ $$$${Find}\:{the}\:{possible}\:{greatest}\:{value}\:{of}\:\mid{x}\:−\:{y}\mid \\ $$ Commented by Yozzia last updated on 03/Sep/16 $$\mathrm{2}^{{x}−{y}}…
Question Number 138613 by byaw last updated on 15/Apr/21 Answered by physicstutes last updated on 15/Apr/21 $$\mathrm{This}\:\mathrm{question}\:\mathrm{is}\:\mathrm{real}\:\mathrm{lol}! \\ $$$$\mathrm{2}.\:\left(\mathrm{a}\right)\:\mathrm{Total}\:\mathrm{length}\:=\:\frac{\mathrm{9}}{\mathrm{4}}\:\mathrm{m}\: \\ $$$$\Rightarrow\:\mathrm{lenght}\:\mathrm{used}\:\mathrm{for}\:\mathrm{a}\:\mathrm{job}\:=\:\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{9}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{m} \\ $$$$\mathrm{length}\:\mathrm{used}\:=\:\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\mathrm{m} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{lenght}\:\mathrm{used}\:\mathrm{for}\:\mathrm{job}\:=\:\frac{\mathrm{1}}{\mathrm{8}}×\frac{\mathrm{3}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{32}}…
Question Number 138600 by Raxreedoroid last updated on 15/Apr/21 $$\mathrm{prove}\:\mathrm{or}\:\mathrm{disprove} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{f}\left({k}\right)={f}\left(\mathrm{1}\right)+\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\left(\frac{\underset{{i}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} {f}\left({i}+\mathrm{1}\right){C}_{{i}−\mathrm{1}} ^{{k}−\mathrm{2}} }{\left({k}−\mathrm{1}\right)!}\:\underset{{i}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\prod}}\left({n}−{i}\right)\right) \\ $$…
Question Number 7519 by Tawakalitu. last updated on 01/Sep/16 $${A}\:\left(\mathrm{2}\:×\:\mathrm{3}\right)\:{rectangle}\:{and}\:{a}\:\left(\mathrm{3}\:×\:\mathrm{4}\right)\:{rectangle}\:{are}\:{contain}\:{within}\: \\ $$$${a}\:{square}\:{without}\:{over}\:{laping}\:{at}\:{any}\:{inferior}\:{point}\:,\:{and}\:{the}\: \\ $$$${sides}\:{of}\:{the}\:{square}\:{are}\:{parallel}\:{to}\:{the}\:{sides}\:{of}\:{the}\:{two}\:{given} \\ $$$${rectangles}.\:{what}\:{is}\:{the}\:{smallest}\:{possible}\:{area}\:{of}\:{the}\:{square}. \\ $$ Commented by Rasheed Soomro last updated on…