MathDB

6

Part of 1992 Poland - Second Round

Problems(1)

x_{n+1} = (x_n+2)/(x_n+1) ,y _{n+1}=(y_n^2+2)/2y_n

Source: Polish MO Recond Round 1992 p6

9/9/2024
The sequences (xn)(x_n) and (yn)(y_n) are defined as follows: x_{n+1} = \frac{x_n+2}{x_n+1},  y_{n+1}=\frac{y_n^2+2}{2y_n}   \text{ for } n= 0,1,2,\ldots. Prove that for every integer n≥0 n\geq 0 the equality yn=x2n−1 y_n = x_{2^n-1} holds.
algebraSequencerecurrence relation