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Let $$a_{n+1} = \frac{a_n + b_n}2,\quad b_{n+1} = \frac{a_{n+1} + b_n}2.$$

Express $a_n$ and $b_n$ in terms of $a_1$, $b_1$ and $n$

Somos
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    And what have you tried? – Rushabh Mehta Jul 15 '19 at 12:25
  • Honestly, I do not know how to express it in English properly as I have studied this topic in Russian. I've got an answer by considering $a_2$, $a_3$, etc in terms of $a_1$ and $b_1$ (and then did the same thing with $b_2$, $b_3$, etc). I noticed some dependencies and then solved a number of recurrence relations. I only want to make sure whether my answers are right or not and discover new solutions – Ilnas Beridze Jul 15 '19 at 12:38
  • thx, it is exactly what I needed :) – Ilnas Beridze Jul 15 '19 at 13:44
  • @Ilnas Beridze:$;$Correcting my earlier comment (now deleted), we can manipulate the given bivariate recurrences to get $$ a_n=\frac{5}{4}a_{n-1}-\frac{1}{4}a_{n-2}\ b_n=\frac{5}{4}b_{n-1}-\frac{1}{4}b_{n-2} $$ for all $n\ge 3$. $$$$ Using standard techniques for solving univariate linear recurrences with constant coefficients, we get $$ a_n = \frac{a_1+2b_1}{3}

    \frac{8(a_1-b_1)}{3} {,\cdot,} \left(\frac{1}{4}\right)^n \ b_n = \frac{a_1+2b_1}{3}

    \frac{4(b_1-a_1)}{3} {,\cdot,} \left(\frac{1}{4}\right)^n $$

    – quasi Jul 15 '19 at 15:29

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Hint: $$\begin{pmatrix}a_{n+1}\\ b_{n+1}\end{pmatrix}=\begin{pmatrix}\frac12&\frac12\\ \frac14&\frac34\end{pmatrix}\begin{pmatrix}a_{n}\\ b_{n}\end{pmatrix} $$