Question Number 118876 by bemath last updated on 20/Oct/20
$${Find}\:{the}\:{value}\:{of}\:\lfloor\:\left(\frac{\mathrm{3}+\sqrt{\mathrm{17}}}{\mathrm{2}}\right)^{\mathrm{6}} \:\rfloor\:. \\ $$
Answered by bobhans last updated on 20/Oct/20
$$\mathrm{2040} \\ $$
Answered by MJS_new last updated on 20/Oct/20
$$\left(\frac{\mathrm{3}+\sqrt{\mathrm{17}}}{\mathrm{2}}\right)^{\mathrm{6}} =\frac{\mathrm{2041}+\mathrm{495}\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{2041}}{\mathrm{2}}+\frac{\mathrm{495}\sqrt{\mathrm{17}}}{\mathrm{2}}−{n}<\mathrm{1}\:\mathrm{with}\:{n}\in\mathbb{N} \\ $$$$\mathrm{2}{n}−\mathrm{2041}\leqslant\mathrm{495}\sqrt{\mathrm{17}}<\mathrm{2}{n}−\mathrm{2039} \\ $$$$\mathrm{all}\:\mathrm{are}\:\mathrm{positive}\:\Rightarrow\:\mathrm{squaring}\:\mathrm{allowed} \\ $$$$\left(\mathrm{2}{n}−\mathrm{2041}\right)^{\mathrm{2}} \leqslant\mathrm{4165435}<\left(\mathrm{2}{n}−\mathrm{2039}\right)^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${n}^{\mathrm{2}} −\mathrm{2041}{n}+\mathrm{64}\leqslant\mathrm{0}<{n}^{\mathrm{2}} −\mathrm{2039}{n}−\mathrm{1976} \\ $$$${n}^{\mathrm{2}} −\mathrm{2041}{n}+\mathrm{64}\leqslant\mathrm{0}\:\Rightarrow\:\frac{\mathrm{2041}−\mathrm{495}\sqrt{\mathrm{17}}}{\mathrm{2}}\leqslant{n}\leqslant\frac{\mathrm{2041}+\mathrm{495}\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} −\mathrm{2039}{n}−\mathrm{1976}>\mathrm{0}\:\Rightarrow\:{n}<\frac{\mathrm{2039}−\mathrm{495}\sqrt{\mathrm{17}}}{\mathrm{2}}\vee{n}>\frac{\mathrm{2039}+\mathrm{495}\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{2039}+\mathrm{495}\sqrt{\mathrm{17}}}{\mathrm{2}}<{n}\leqslant\frac{\mathrm{2041}+\mathrm{495}\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$$\mathrm{2039}.\mathrm{96}...<{n}\leqslant\mathrm{2040}.\mathrm{96}... \\ $$$$\mathrm{but}\:{n}\in\mathbb{N} \\ $$$$\Rightarrow\:{n}=\mathrm{2040} \\ $$
Answered by TANMAY PANACEA last updated on 20/Oct/20
$${y}=\sqrt{{x}}\: \\ $$$$\frac{{dy}}{{dx}}\approx\frac{\bigtriangleup{y}}{\bigtriangleup{x}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\rightarrow\bigtriangleup{y}=\frac{\bigtriangleup{x}}{\mathrm{2}\sqrt{{x}}} \\ $$$${when}\:{x}=\mathrm{16}\:\:\:{y}=\sqrt{\mathrm{16}}\:=\mathrm{4} \\ $$$${x}+\bigtriangleup{x}=\mathrm{17}\:\:\:{y}+\bigtriangleup{y}=? \\ $$$$\bigtriangleup{y}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{16}}\:}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\sqrt{\mathrm{17}}\:=\mathrm{4}+\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{33}}{\mathrm{8}} \\ $$$$\left(\frac{\mathrm{3}+\frac{\mathrm{33}}{\mathrm{8}}}{\mathrm{2}}\right)^{\mathrm{6}} =\left(\frac{\mathrm{3}+\mathrm{4}.\mathrm{125}}{\mathrm{2}}\right)^{\mathrm{6}} =\left(\mathrm{1}.\mathrm{5}+\mathrm{2}.\mathrm{0625}\right)^{\mathrm{6}} =\left(\mathrm{3}.\mathrm{5625}\right)^{\mathrm{6}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$