Difference between revisions of "Mock AIME 2 Pre 2005 Problems/Problem 2"

(Created page with "== Problem == <math>x</math> is a real number with the property that <math>x+\tfrac1x = 3</math>. Let <math>S_m = x^m + \tfrac{1}{x^m}</math>. Determine the value of <math>S_7...")
 
(Solution)
 
Line 4: Line 4:
 
== Solution ==
 
== Solution ==
 
We can calculate <cmath>x^2 + \dfrac{1}{x^2} = \left(x + \dfrac{1}{x}\right)^2 - 2 = 3^2 -2 = 7.</cmath> Similarly, <cmath>x^3 + \dfrac{1}{x^3} = \left(x + \dfrac{1}{x}\right) \left(x^2 + \dfrac{1}{x^2}\right) - \left(x + \dfrac{1}{x}\right) = 3 \cdot 7 - 3 = 18</cmath> and <cmath>x^4 + \dfrac{1}{x^4} = \left(x^2 + \dfrac{1}{x^2}\right)^2 - 2 = 7^2 - 2 = 47.</cmath> Finally, <cmath>x^7 + \dfrac{1}{x^7} = \left(x^3 + \dfrac{1}{x^3}\right) \left(x^4 + \dfrac{1}{x^4}\right) - \left(x + \dfrac{1}{x}\right) = 18 \cdot 47 - 3 = \boxed{843}.</cmath>
 
We can calculate <cmath>x^2 + \dfrac{1}{x^2} = \left(x + \dfrac{1}{x}\right)^2 - 2 = 3^2 -2 = 7.</cmath> Similarly, <cmath>x^3 + \dfrac{1}{x^3} = \left(x + \dfrac{1}{x}\right) \left(x^2 + \dfrac{1}{x^2}\right) - \left(x + \dfrac{1}{x}\right) = 3 \cdot 7 - 3 = 18</cmath> and <cmath>x^4 + \dfrac{1}{x^4} = \left(x^2 + \dfrac{1}{x^2}\right)^2 - 2 = 7^2 - 2 = 47.</cmath> Finally, <cmath>x^7 + \dfrac{1}{x^7} = \left(x^3 + \dfrac{1}{x^3}\right) \left(x^4 + \dfrac{1}{x^4}\right) - \left(x + \dfrac{1}{x}\right) = 18 \cdot 47 - 3 = \boxed{843}.</cmath>
 +
 +
-MP8148
  
 
== See also ==
 
== See also ==

Latest revision as of 15:14, 4 August 2019

Problem

$x$ is a real number with the property that $x+\tfrac1x = 3$. Let $S_m = x^m + \tfrac{1}{x^m}$. Determine the value of $S_7$.

Solution

We can calculate \[x^2 + \dfrac{1}{x^2} = \left(x + \dfrac{1}{x}\right)^2 - 2 = 3^2 -2 = 7.\] Similarly, \[x^3 + \dfrac{1}{x^3} = \left(x + \dfrac{1}{x}\right) \left(x^2 + \dfrac{1}{x^2}\right) - \left(x + \dfrac{1}{x}\right) = 3 \cdot 7 - 3 = 18\] and \[x^4 + \dfrac{1}{x^4} = \left(x^2 + \dfrac{1}{x^2}\right)^2 - 2 = 7^2 - 2 = 47.\] Finally, \[x^7 + \dfrac{1}{x^7} = \left(x^3 + \dfrac{1}{x^3}\right) \left(x^4 + \dfrac{1}{x^4}\right) - \left(x + \dfrac{1}{x}\right) = 18 \cdot 47 - 3 = \boxed{843}.\]

-MP8148

See also

Mock AIME 2 Pre 2005 (Problems, Source)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15