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Category: Number Theory

Find-2020-term-from-series-1-1-2-1-1-2-3-1-2-2-1-3-4-1-3-2-2-3-1-4-5-1-4-2-is-A-2019-2020-B-61-4-C-63-1-D-96-4-E-2020-2019

Question Number 101418 by john santu last updated on 02/Jul/20 $$\mathrm{Find}\:\mathrm{2020}\:\mathrm{term}\:\mathrm{from}\:\mathrm{series} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}},\frac{\mathrm{2}}{\mathrm{1}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{1}},\frac{\mathrm{2}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{4}}{\mathrm{1}},\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$,\frac{\mathrm{5}}{\mathrm{1}},\frac{\mathrm{4}}{\mathrm{2}},…\:\mathrm{is}\:\_\_\_ \\ $$$$\left(\mathrm{A}\right)\:\frac{\mathrm{2019}}{\mathrm{2020}}\:\:\:\:\:\:\left(\mathrm{B}\right)\:\frac{\mathrm{61}}{\mathrm{4}}\:\:\:\:\:\left(\mathrm{C}\right)\frac{\mathrm{63}}{\mathrm{1}} \\ $$$$\left(\mathrm{D}\right)\:\frac{\mathrm{96}}{\mathrm{4}}\:\:\:\:\:\:\left(\mathrm{E}\right)\:\frac{\mathrm{2020}}{\mathrm{2019}} \\ $$ Commented by PRITHWISH SEN…

Question-35872

Question Number 35872 by Rasheed.Sindhi last updated on 25/May/18 Commented by Tinkutara last updated on 25/May/18 Yes this was also my doubt! Answered by Rasheed.Sindhi last updated on 26/May/18 $$\mathrm{Note}:\:\mathrm{Capital}\:\mathrm{C}\:\mathrm{and}\:\mathrm{small}\:\mathrm{c}\:\mathrm{has}\:\mathrm{been}…

Question-166911

Question Number 166911 by rexford last updated on 02/Mar/22 Answered by MathsFan last updated on 02/Mar/22 $$\boldsymbol{{suppose}}\:\sqrt{\mathrm{2}}\:\boldsymbol{{is}}\:\boldsymbol{{rational}} \\ $$$$\sqrt{\mathrm{2}}=\frac{\boldsymbol{{p}}}{\boldsymbol{{q}}}\:\Rightarrow\:\mathrm{2}=\frac{\boldsymbol{{p}}^{\mathrm{2}} }{\boldsymbol{{q}}^{\mathrm{2}} }\:\Rightarrow\:\:\mathrm{2}\boldsymbol{{q}}^{\mathrm{2}} =\boldsymbol{{p}}^{\mathrm{2}} …..\left(\mathrm{1}\right) \\ $$$$\:\mathrm{2}\:\boldsymbol{{divides}}\:\boldsymbol{{p}}…

1-1-2-11-1-3-11-1-4-11-1-1-2-11-1-3-11-1-4-11-

Question Number 101043 by bobhans last updated on 30/Jun/20 $$\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{11}} }\:+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{11}} }\:+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{11}} }\:+\:…}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{11}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{11}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{11}} }+…}\:=? \\ $$ Answered by john santu last updated on…

Question-166530

Question Number 166530 by amin96 last updated on 21/Feb/22 Answered by qaz last updated on 22/Feb/22 $$\mathrm{6}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2n}\right)!}{\left(\mathrm{n}!\right)^{\mathrm{2}} \mathrm{2}^{\mathrm{4n}+\mathrm{1}} \left(\mathrm{2n}+\mathrm{1}\right)} \\ $$$$=\mathrm{3}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2n}}\\{\mathrm{n}}\end{pmatrix}\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2n}}…