There is a well-known estimate for the sum of all binomial coefficients $\binom{n}{k}$ satisfying $k \leq \alpha n$ for some $\alpha$ satisfying $0 < \alpha \leq 1/2$:


$$ \sum\_{k=0}^{\alpha n}\binom{n}{k} = 2^{(H(\alpha) + o(1))n}$$


where $H(\alpha) = -\alpha\log\_2(\alpha) - (1-\alpha)\log\_2(1-\alpha)$ is the binary entropy function. 


My question is whether there exists a similar estimate when we weight the $k$-th binomial coefficient by $\lambda^k$ for some $\lambda > 0$. That is, I would like to estimate the following sum:


$$ \sum\_{k=0}^{\alpha n} \binom{n}{k} \lambda^k $$



Estimating a partial sum of weighted binomial coefficients
Expanding on the previous answers. I'm taking $\lambda$ and $\alpha$ to be constants which do not vary as $n\to\infty$. 


If $α<λ/(λ+1)$ then the sum is within a constant of the last term. In fact the largest terms are approximately in geometric progression so you can get it quite accurately by computing the ratio.


If $α>λ/(λ+1)$, almost all of the complete binomial expansion is present, so the sum equals $(1+o(1))(1+λ)^n$.


If $α≈p$, then Russell's normal approximation will be good. (This needs some work to clarify whether the geometric approximation of the lower tail is good right up to the point where the normal approximation begins to be good. I think it is.)


