Problem
ALG-B1-M05-P010 Decreasing Difference of Roots
#10
★★★☆☆ Level 3 of 5
Let \(1
Set \(A=\sqrt[2^{n+1}]{a}\), \(B=\sqrt[2^{n+1}]{b}\).
Then \(B>A>1\), \(x_{n+1}=2^{n+1}(B-A)\), and \(x_n=2^n(B^2-A^2)=2^{n+1}(B-A)\frac{A+B}{2}\).
Since \(\frac{A+B}{2}>1\), we get \(x_n>x_{n+1}\).
This trick reappears in a classified rewrite.