Question Number 94690 by abony1303 last updated on 20/May/20
Commented by abony1303 last updated on 20/May/20
$${Pls}\:{Help}! \\ $$
Commented by prakash jain last updated on 20/May/20
$$ \\ $$$$\mathrm{First}\:\mathrm{term}\:{a}_{\mathrm{1}} =\mathrm{1} \\ $$$$\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} =\mathrm{2} \\ $$$${a}_{\mathrm{4}} ,{a}_{\mathrm{5}} ,{a}_{\mathrm{6}} =\mathrm{3} \\ $$$${a}_{{k}} ={n}\:\mathrm{for}\:\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}<{k}\leqslant\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${k}=\mathrm{2019} \\ $$$${n}=\mathrm{64} \\ $$$$\frac{\mathrm{64}×\mathrm{63}}{\mathrm{2}}=\mathrm{2016}<\mathrm{2019}\leqslant\frac{\mathrm{64}×\mathrm{65}}{\mathrm{2}}=\mathrm{2080} \\ $$
Commented by abony1303 last updated on 20/May/20
$${where}\:{can}\:{I}\:{learn}\:{this}\:{kind}\:{of}\:{questions} \\ $$$${I}\:{mean}\:{what}'{s}\:{the}\:{name}\:{of}\:{this}\:{theme}? \\ $$
Commented by mr W last updated on 21/May/20
$${a}_{{k}} =\lceil\frac{\sqrt{\mathrm{8}{k}+\mathrm{1}}−\mathrm{1}}{\mathrm{2}}\rceil \\ $$$${a}_{\mathrm{2019}} =\lceil\frac{\sqrt{\mathrm{8}×\mathrm{2019}+\mathrm{1}}−\mathrm{1}}{\mathrm{2}}\rceil=\mathrm{64} \\ $$