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Author: Tinku Tara

A-1-2-2-4-3-6-14-28-B-1-3-2-3-14-29-B-

Question Number 10340 by konen last updated on 04/Feb/17 $$\mathrm{A}=\mathrm{1}×\mathrm{2}\:+\mathrm{2}×\mathrm{4}\:+\mathrm{3}×\mathrm{6}+…+\mathrm{14}×\mathrm{28} \\ $$$$\mathrm{B}=\mathrm{1}×\mathrm{3}\:+\mathrm{2}×\mathrm{3}\:+…+\mathrm{14}×\mathrm{29} \\ $$$$\Rightarrow\mathrm{B}=? \\ $$ Commented by mrW1 last updated on 04/Feb/17 $${the}\:{definition}\:{of}\:{B}\:{is}\:{not}\:{clear}. \\…

Prove-that-n-N-n-1-n-lnt-dt-ln-n-1-2-

Question Number 141409 by Willson last updated on 18/May/21 $$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\forall\mathrm{n}\in\mathbb{N}\:\:\:\:\underset{\mathrm{n}} {\int}^{\:\mathrm{n}+\mathrm{1}} \mathrm{lnt}\:\mathrm{dt}\:\leqslant\:\mathrm{ln}\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$ Answered by TheSupreme last updated on 19/May/21 $$\int{ln}\left({t}\right)={xln}\left({t}\right)−{x} \\…

A-1-2-2-3-3-4-10-11-B-3-8-6-12-9-16-30-44-A-B-

Question Number 10339 by konen last updated on 04/Feb/17 $$\mathrm{A}=\mathrm{1}×\mathrm{2}\:+\:\mathrm{2}×\mathrm{3}\:+\mathrm{3}×\mathrm{4}+…+\mathrm{10}×\mathrm{11} \\ $$$$\mathrm{B}=\mathrm{3}×\mathrm{8}\:+\mathrm{6}×\mathrm{12}\:+\mathrm{9}×\mathrm{16}+…+\mathrm{30}×\mathrm{44} \\ $$$$\Rightarrow\frac{\mathrm{A}}{\mathrm{B}}=? \\ $$ Answered by mrW1 last updated on 04/Feb/17 $${a}_{{n}} ={n}×\left({n}+\mathrm{1}\right)…

Find-the-minimum-value-of-k-such-that-for-arbitrary-a-b-gt-0-we-have-a-1-3-b-1-3-k-a-b-1-3-

Question Number 141405 by iloveisrael last updated on 18/May/21 $$\:\:\:\:\:{Find}\:{the}\:{minimum}\:{value}\:{of}\:{k} \\ $$$$\:\:\:\:\:{such}\:{that}\:{for}\:{arbitrary}\:{a},{b}\:>\mathrm{0} \\ $$$$\:\:\:\:\:{we}\:{have}\:\:\sqrt[{\mathrm{3}\:}]{{a}}\:+\:\sqrt[{\mathrm{3}\:}]{{b}}\:\leqslant\:{k}\:\sqrt[{\mathrm{3}\:}]{{a}+{b}}\: \\ $$ Answered by EDWIN88 last updated on 18/May/21 $$\:\mathrm{consider}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{x}} \\…

fjnd-inverse-of-matrix-2-5-1-3-

Question Number 141400 by Raffaqet last updated on 18/May/21 $${fjnd}\:{inverse}\:{of}\:{matrix}\begin{bmatrix}{\mathrm{2}}&{−\mathrm{5}}\\{\mathrm{1}}&{\mathrm{3}}\end{bmatrix} \\ $$ Answered by iloveisrael last updated on 18/May/21 $$\:{A}^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{6}−\left(−\mathrm{5}\right)}\:\begin{bmatrix}{\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{5}}\\{−\mathrm{1}\:\:\:\:\:\mathrm{2}}\end{bmatrix} \\ $$$${A}^{−\mathrm{1}} \:=\:\begin{bmatrix}{\mathrm{3}/\mathrm{11}\:\:\:\:\:\:\:\:\:\mathrm{5}/\mathrm{11}}\\{−\mathrm{1}/\mathrm{11}\:\:\:\:\:\mathrm{2}/\mathrm{11}}\end{bmatrix} \\…