Let $\xi_1, \xi_2,...,\xi_n$ be independent Bernoulli random variables in $(\Omega,\mathcal{P}(\Omega),\mathbb{P})$ and $$\mathbb{P}(\xi_i=0)=1-\lambda_i\Delta$$ and $$\mathbb{P}(\xi_i=1)=\lambda_i\Delta.$$ Here $\lambda_1,\lambda_2,...,\lambda_n$ are positive parameters and $0< \Delta <\frac{1}{2\max\{\lambda_1,\lambda_2,...,\lambda_n\}}$.
We need to show $$\mathbb{P}(\xi_1+ \xi_2+...+\xi_n=1)=(\sum_{i=1}^{n}\lambda_i)\Delta + \mathcal{R}\Delta^2,$$ where $|\mathcal{R}| \leq 2(\sum_{i=1}^{n}\lambda_i^2+(\sum_{i=1}^{n}\lambda_i)^2)$
How I should start and what to do? I really have no idea. Can somebody explain this to me?
We have \begin{align*} P\left(\sum_{i=1}^{n}\xi_i=1\right) &= \sum_{i=1}^{n}P(\xi_i=1, \{\xi_j = 0: j\neq i\}) \\ &=\sum_{i=1}^n\lambda_i\Delta\left(\prod_{j\neq i}(1-\lambda_j \Delta)\right) \\ &= \sum_{i=1}^n\lambda_i\Delta\left(1 - \Delta \sum_{j\neq i}\lambda_j + \Delta^2\sum_{j_1, j_2\neq i}\lambda_{j_1}\lambda_{j_2} - \cdots\right) \\ &= \sum_{i=1}^n\lambda_i\Delta - \Delta^2 \left\{2\sum_{i=1}^{n}\sum_{j = 1}^{i-1}\lambda_i\lambda_j\right\} + \cdots \end{align*} Note that the term in curly brackets equals \begin{align*} \left(\sum_{i=1}^{n}\lambda_i\right)^2 - \sum_{i=1}^{n}\lambda_i^2 \end{align*} which is clearly less in absolute value than $2(\sum_{i=1}^{n}\lambda_i^2 + \left(\sum_{i=1}^{n}\lambda_i\right)^2)$. Since \begin{align*} 0 < \Delta < \frac{1}{2\max(\lambda_1, \cdots, \lambda_n)} \end{align*} The alternating series of sums are decreasing, so therefore the largest the sum can be is less than the remainder bound.