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 2 last updated on 02/Jul/20

2020=((4040)/2) = ((4032)/2) +4=((64×63)/2) + 4 =(((61+4−1)(61+4−2))/2)+4 i.e the term is ((61)/4) ans(B) please check sorry it was a typo

$$\mathrm{2020}=\frac{\mathrm{4040}}{\mathrm{2}}\:=\:\frac{\mathrm{4032}}{\mathrm{2}}\:+\mathrm{4}=\frac{\mathrm{64}×\mathrm{63}}{\mathrm{2}}\:+\:\mathrm{4} \\ $$$$=\frac{\left(\mathrm{61}+\mathrm{4}−\mathrm{1}\right)\left(\mathrm{61}+\mathrm{4}−\mathrm{2}\right)}{\mathrm{2}}+\mathrm{4} \\ $$$$\boldsymbol{\mathrm{i}}.\boldsymbol{\mathrm{e}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{is}}\:\frac{\mathrm{61}}{\mathrm{4}}\:\boldsymbol{\mathrm{ans}}\left(\boldsymbol{\mathrm{B}}\right)\:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$$$\mathrm{sorry}\:\mathrm{it}\:\mathrm{was}\:\mathrm{a}\:\mathrm{typo} \\ $$

Commented by bemath last updated on 02/Jul/20