How does $\sum_{t=0}^\infty(1-\frac2n)^t\frac{e^{-n\lambda }(n\lambda)^t}{t!}=e^{-n\lambda}\sum_{t=0}^\infty \frac{[\lambda(n-2)]^t}{t!}.$

Question

How does $$\sum_{t=0}^\infty(1-\frac2n)^t\frac{e^{-n\lambda }(n\lambda)^t}{t!}=e^{-n\lambda}\sum_{t=0}^\infty \frac{[\lambda(n-2)]^t}{t!}.$$All I see is $e^{-n\lambda}$ getting pulled out.

score 1 · Accepted Answer · answered Mar 22 '16 at 07:05

1

$$ \left(1-\frac2n\right)^t\frac{(n\lambda)^t}{t!} = \left(\frac{n-2}n\right)^t \frac{(n\lambda)^t}{t!}= \frac{[\lambda(n-2)]^t}{t!}. $$

answered Mar 22 '16 at 07:05

Martín-Blas Pérez Pinilla

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How does $\sum_{t=0}^\infty(1-\frac2n)^t\frac{e^{-n\lambda }(n\lambda)^t}{t!}=e^{-n\lambda}\sum_{t=0}^\infty \frac{[\lambda(n-2)]^t}{t!}.$

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