Question Number 75739 by Master last updated on 16/Dec/19
Commented by Master last updated on 16/Dec/19
![[ x ]−whole part](Q75740.png)
$$\left[\:\mathrm{x}\:\right]−\mathrm{whole}\:\mathrm{part} \\ $$
Answered by mr W last updated on 16/Dec/19
![x+(1/x)=(7/3) x^2 −(7/3)x+1=0 x=((7±(√(13)))/6) p_n =x^n +(1/x^n )=(((7+(√(13)))/6))^n +(((7−(√(13)))/6))^n p_(13) =x^(13) +(1/x^(13) )=(((7+(√(13)))/6))^(13) +(((7−(√(13)))/6))^(13) [x^(13) +(1/x^(13) )]=[(((7+(√(13)))/6))^(13) +(((7−(√(13)))/6))^(13) ] =[((2621455207)/(1594323))]=1644](Q75754.png)
$${x}+\frac{\mathrm{1}}{{x}}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{7}}{\mathrm{3}}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{7}\pm\sqrt{\mathrm{13}}}{\mathrm{6}} \\ $$$${p}_{{n}} ={x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=\left(\frac{\mathrm{7}+\sqrt{\mathrm{13}}}{\mathrm{6}}\right)^{{n}} +\left(\frac{\mathrm{7}−\sqrt{\mathrm{13}}}{\mathrm{6}}\right)^{{n}} \\ $$$${p}_{\mathrm{13}} ={x}^{\mathrm{13}} +\frac{\mathrm{1}}{{x}^{\mathrm{13}} }=\left(\frac{\mathrm{7}+\sqrt{\mathrm{13}}}{\mathrm{6}}\right)^{\mathrm{13}} +\left(\frac{\mathrm{7}−\sqrt{\mathrm{13}}}{\mathrm{6}}\right)^{\mathrm{13}} \\ $$$$\left[{x}^{\mathrm{13}} +\frac{\mathrm{1}}{{x}^{\mathrm{13}} }\right]=\left[\left(\frac{\mathrm{7}+\sqrt{\mathrm{13}}}{\mathrm{6}}\right)^{\mathrm{13}} +\left(\frac{\mathrm{7}−\sqrt{\mathrm{13}}}{\mathrm{6}}\right)^{\mathrm{13}} \right] \\ $$$$=\left[\frac{\mathrm{2621455207}}{\mathrm{1594323}}\right]=\mathrm{1644} \\ $$
Commented by Master last updated on 16/Dec/19
$$\mathrm{great}\:! \\ $$