Your sum can also be thought of as the first $\alpha n$ terms in a binomial distribution with probability of success $p=1-\frac1{\lambda+1}$. So, it is closely approximated by a normal distribution with mean $np$ and standard deviation , $\sqrt{np(1-p)}$, i.e., $$\sum_{k=0}^{\alpha n} \binom nk \lambda^k\approx (1-p)^{-n}\Phi\left((\alpha-p)\sqrt{\frac n{p(1-p)}}\right),$$ where $\Phi$ is the cumulative standard normal distribution.
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Your sum can also be thought of as the first $\alpha n$ terms in a binomial distribution with probability of success $p=1-\frac1{\lambda+1}$. So, it is closely approximated by a normal distribution with mean $np$ and standard deviation, i.e., $$\sum_{k=0}^{\alpha n} \binom nk \lambda^k\approx (1-p)^{-n}\Phi\left((\alpha-p)\sqrt{\frac n{p(1-p)}}\right),$$ where $\Phi$ is the cumulative standard normal distribution. |
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