Suppose we are given $(\lambda a + \bar{\lambda}b+O(\lambda^2))^{n}$, where $0 < \lambda < 1$ and $\bar{\lambda} := 1-\lambda$; also, $0 < a,b < 1$. $O(\cdot)$ is the traditional Big-Oh notation.
The question is that assume we want $n$ to grow large, under which condition the $O(\lambda^2)$ term becomes negligible, so that we might approximate $(\lambda a + \bar{\lambda}b+O(\lambda^2))^{n}$ as $(\lambda a + \bar{\lambda} b)^n$ ?
The usual thing to do here is factor out the term that's large to form a ratio. Let $u:=(\lambda a+\bar{\lambda}b)$. Then
$$(u+O(\lambda^2))^n=u^n\left(1+\frac{O(\lambda^2)}{u}\right)^n.$$
Now,
$$(u+O(\lambda^2))^n=\exp(\log((u+O(\lambda^2))^n)=u^n\exp\left(\frac{O(\lambda^2)}{u}+O\left(\frac{\lambda^2}{u^2n}\right)\right),$$
so if $n\gg \lambda^2/u^2$, you have a decent estimate, assuming $O(\lambda^2)/u$ is small, which by the looks of it is $O(\lambda^2)$ if $\lambda$ is small.