Question Number 138842 by peter frank last updated on 18/Apr/21 $${Find}\:{the}\:{sum}\:{to}\:{infinity} \\ $$$$\mathrm{1}+\frac{\mathrm{3}{x}}{\mathrm{1}!}+\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{7}{x}^{\mathrm{3}} }{\mathrm{3}!}+.. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 7750 by Tawakalitu. last updated on 13/Sep/16 $${Let}\:{n}\:=\:\left(\mathrm{2}^{\mathrm{31}} \right)\:×\:\left(\mathrm{3}^{\mathrm{19}} \right)\:{how}\:{many}\:{positive}\:{integer} \\ $$$${divisors}\:{of}\:{n}^{\mathrm{2}} \:{are}\:{less}\:{than}\:{n}\:{but}\:{do}\:{not}\:{divide}\:{n} \\ $$ Commented by Yozzia last updated on 13/Sep/16 $${n}^{\mathrm{2}}…
Question Number 7743 by Tawakalitu. last updated on 13/Sep/16 $${All}\:{the}\:{terms}\:{of}\:{the}\:{arithmetic}\:{progession}\: \\ $$$${u}_{\mathrm{1}} ,\:{u}_{\mathrm{2}} ,\:{u}_{\mathrm{3}} ,\:…\:{u}_{{n}} \:\:{are}\:{positive}\:.\:{use}\:{induction}\:{to} \\ $$$${prove}\:{that}\:{for}\:{n}\:\geqslant\:\mathrm{2} \\ $$$$\frac{\mathrm{1}}{{u}_{\mathrm{1}} {u}_{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{u}_{\mathrm{2}} {u}_{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{u}_{\mathrm{3}} {u}_{\mathrm{4}}…
Question Number 138789 by 676597498 last updated on 18/Apr/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 7723 by FilupSmith last updated on 12/Sep/16 $${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}}…