I'm new to recurrence relations, and I'm inventing some exercises by myself in order to solve them. Considering this recurrence relation
$$ x_n = x_{n-1} - \lambda (x_{n-1} - 1)^2 $$
where $\lambda \in (0,1)$. Does this recurrence relationship has a solution for a generic point $x_1 = a>1$?
This is not an answer to the original question (an explicit formula for $x_n$) but to the request of the OP about the the behavior of the sequence $\{x_n\}$, and is too long for a comment.
Let $\lambda>0$ and $f(x)=x-\lambda(x-1)^2$. We have $f(1)=f(1+1/\lambda)=1$. If $a=1$ or $a=1+1/\lambda$, then $x_n=1$ for all $n>1$.
It is easy to check that $1<f(x)<x$ if $1<x<1+1/\lambda$. This implies that if $1<a<1+1/\lambda$, then $\{x_n\}$ is strictly decreasing and bounded below by $1$. Thus $\{x_n\}$ converges, and the only possible limit is $1$.
It is also easy to see that if $a\notin[1,1+1/\lambda]$, then $x_n\to-\infty$.