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Category: Algebra

Question-203891

Question Number 203891 by York12 last updated on 02/Feb/24 $$ \\ $$ Answered by sniper237 last updated on 01/Feb/24 $${The}\:{last}\:{factor}\:{should}\:{be}\:\left(\frac{{c}}{{a}+{b}}+\frac{{a}}{{b}+{c}}\right) \\ $$$${If}\:\:{so}\:,\:{Let}\:{named}\:{P}\:{that}\:{product} \\ $$$${Divide}\:{each}\:{factor}\:{by}\:{c},{a},{b}\:{resp} \\…

Evaluate-26-15-3-1-3-26-15-3-1-3-

Question Number 203826 by necx122 last updated on 29/Jan/24 $${Evaluate} \\ $$$$\left(\mathrm{26}\:+\:\mathrm{15}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{3}} +\:\left(\mathrm{26}\:−\:\mathrm{15}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$ Answered by AST last updated on 29/Jan/24 $${Let}\:{a}=\sqrt[{\mathrm{3}}]{\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}};{b}=\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}}\Rightarrow{ab}=\mathrm{1} \\ $$$${a}^{\mathrm{3}}…

determine-whether-the-series-is-convergent-or-divergent-n-1-n-4n-2-1-

Question Number 203774 by Calculusboy last updated on 27/Jan/24 $$\boldsymbol{{determine}}\:\boldsymbol{{whether}}\:\boldsymbol{{the}}\:\boldsymbol{{series}}\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{convergent}}\:\boldsymbol{{or}}\:\boldsymbol{{divergent}} \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\boldsymbol{{n}}}{\:\sqrt{\mathrm{4}\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{1}}} \\ $$ Answered by witcher3 last updated on…

Question-203742

Question Number 203742 by Calculusboy last updated on 27/Jan/24 Answered by mr W last updated on 27/Jan/24 $${x}^{\mathrm{2024}} +{x}^{\mathrm{2024}} −\mathrm{2024}×\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2023}} +…=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2024}} −\mathrm{506}{x}^{\mathrm{2023}} +…=\mathrm{0}…

Question-203771

Question Number 203771 by Calculusboy last updated on 27/Jan/24 Answered by DwaipayanShikari last updated on 27/Jan/24 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\begin{pmatrix}{{n}+\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}!}{\left({n}+\mathrm{3}\right)!\mathrm{3}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}!}\underset{{n}=\mathrm{0}}…