Menu Close

Author: Tinku Tara

Question-127931

Question Number 127931 by A8;15: last updated on 03/Jan/21 Commented by talminator2856791 last updated on 04/Jan/21 $$\: \\ $$$$\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{series}\:\mathrm{for}\:\mathrm{which}\:\mathrm{you}\:\mathrm{want}\:\mathrm{evaluate}? \\ $$ Commented by A8;15: last…

Prove-that-if-the-lengths-of-a-triangle-form-an-arithmetic-progression-then-the-centre-of-incircle-and-the-centroid-of-triangle-lie-on-a-line-parallel-to-the-side-of-middle-length-of-the-triangle-

Question Number 62380 by ajfour last updated on 20/Jun/19 $${Prove}\:{that}\:{if}\:{the}\:{lengths}\:{of}\:{a}\: \\ $$$${triangle}\:{form}\:{an}\:{arithmetic} \\ $$$${progression},\:{then}\:{the}\:{centre}\:{of} \\ $$$${incircle}\:{and}\:{the}\:{centroid}\:{of} \\ $$$${triangle}\:{lie}\:{on}\:{a}\:{line}\:{parallel}\:{to} \\ $$$${the}\:{side}\:{of}\:{middle}\:{length}\:{of}\:{the} \\ $$$${triangle}. \\ $$ Answered…

Dividing-by-the-factor-of-divisor-find-the-quotient-and-remainder-of-715-12-

Question Number 127912 by PRITHWISH SEN 2 last updated on 03/Jan/21 $$\mathrm{Dividing}\:\mathrm{by}\:\mathrm{the}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{divisor}\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\mathrm{quotient}\:\mathrm{and}\:\mathrm{remainder}\:\mathrm{of}\:\:\mathrm{715}\boldsymbol{\div}\mathrm{12} \\ $$ Answered by physicstutes last updated on 03/Jan/21 $$\:\mathrm{715}\:=\:\mathrm{12}\left(\mathrm{59}\right)\:+\:\mathrm{7} \\…

1-1-3-2-1-3-5-3-1-3-5-7-4-1-3-5-7-9-

Question Number 127908 by bramlexs22 last updated on 03/Jan/21 $$\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}+\frac{\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}}+…=? \\ $$ Answered by liberty last updated on 03/Jan/21 $$\:\mathrm{Let}\:\lambda\:=\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{n}}{\mathrm{1}.\mathrm{3}.\mathrm{5}…\left(\mathrm{2n}+\mathrm{1}\right)}\: \\ $$$$\mathrm{consider}\:\frac{\mathrm{n}}{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2n}+\mathrm{1}\right)}\:=\:\frac{\mathrm{n}}{\mathrm{1}.\left(\mathrm{3}.\mathrm{5}…\left(\mathrm{2n}−\mathrm{1}\right)\right).\left(\mathrm{2n}+\mathrm{1}\right)} \\…