Proving $mathbb{P}(xi_1+ xi_2+…+xi_n=1)=(sum_{i=1}^{n}lambda_i)Delta + mathcal{R}Delta^2$
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Let $xi_1, xi_2,...,xi_n$ be independent Bernoulli random variables in $(Omega,mathcal{P}(Omega),mathbb{P})$ and $$mathbb{P}(xi_i=0)=1-lambda_iDelta$$ and $$mathbb{P}(xi_i=1)=lambda_iDelta.$$ Here $lambda_1,lambda_2,...,lambda_n$ are positive parameters and $0< Delta <frac{1}{2max{lambda_1,lambda_2,...,lambda_n}}$ . 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?
probability probability-theory random-variables bernoulli-numbers
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