Looshim A.L.

The sequence $x_{0}, x_{1}, x_{2}, \ldots$ is defined by the following conditions: $x_{0}=1, x_{1}=\lambda$, for any $n>1$ the equality holds

$$
(\alpha+\beta)^{n} x_{n}=\alpha^{n} x_{n} x_{0}+\alpha^{n-1} \beta x_{n-1} x_{1}+\alpha^{n-2} \beta^{2} x_{n-2} x_{2}+\ldots+\beta^{n} x_{0} x_{n} .
$$

Here $\alpha, \beta, \lambda$ are given positive numbers. Find $x_{n}$ and determine for which $n$ the value of $x_{n}$ is the largest.
We will prove by induction that $x_{n}=\frac{\lambda^{n}}{n!}$. The base case $(n=0,1)$ is given in the condition.

Induction step. Suppose $x_{k}=\frac{\lambda^{k}}{k!}$ for all $k \leq n-1$. Then from the equalities

$$
\left((\alpha+\beta)^{n}-\alpha^{n}-\beta^{n}\right) x_{n}=\sum_{k=1}^{n-1} \frac{\alpha^{n-k} \beta^{k} \lambda^{n}}{(n-k)!k!}=\sum_{k=1}^{n-1} \frac{\alpha^{n-k} \beta^{k} \lambda^{n}}{(n-k)!k!}=\left((\alpha+\beta)^{n}-\alpha^{n}-\beta^{n}\right) \cdot \frac{\lambda^{n}}{n!}
$$

it follows that $x_{n}=\frac{\lambda^{n}}{n!}$, since the common factor on both sides of the equation is positive for $n>1$.

The inequality $\frac{\lambda^{n-1}}{(n-1)!} \leq \frac{\lambda^{n}}{n!}$ is equivalent to the inequality $n \leq \lambda$. Thus, when transitioning from $n-1$ to $n$, the number $x_{n}$ increases if

$n \leq \lambda$. Therefore, the maximum value of $x_{n}$ is achieved at $n=[\lambda]$; if $\lambda$ is an integer, then $x_{\lambda-1}=x_{\lambda}$ are the two largest numbers in the sequence $x_{n}$.

## Answer

$x_{n}=\frac{\lambda^{n}}{n!} ;$ at $n=[\lambda]$