Using independence, EetΣXi=EΠ\PiΠ etXi=Π\PiΠEetXi=(pet+(1-p))n,
where the Xi are independent Bernoulli random variables. Equivalently
∑k=0netk(kn)pk(1−p)n−k=∑k=0n(kn)(pet)k(1−p)n−k=(pet+(1−p)n)\sum^n_{k=0}e^{tk}(^n_k)p^k(1-p)^{n-k}=\sum^n_{k=0}(^n_k)(pe^{t})^k(1-p)^{n-k}=(pe^t+(1-p)^n)∑k=0netk(kn)pk(1−p)n−k=∑k=0n(kn)(pet)k(1−p)n−k=(pet+(1−p)n) by the Binomial formula.
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