An insurance company pays \$10000 for the first loss, \$7500 for the second loss and \$5000 for each successive loss. If the number of losses follows a Poisson distribution with mean 2.5 then what is the expected payment due to losses.
I viewed the problem as follows: If there is 1 loss then there is a payment of \$10000, if there are 2 losses then there is a payment of \$17500 (payment for first loss plus payment for second loss), and for any $k \ge 3$ if there are $k$ losses then there is a payment of \$17500+5000$k$.
If $X$ denotes the number of losses then $P(K=k)=\dfrac{e^{-2.5}2.5^k}{k!}$ and the expected payment would be
$$ 10000P(X=1)+17500P(X=2)+\sum_{k=3}^\infty (17500+5000k)P(X=k) $$
But when I calculate this I get 23433.2 and the correct answer needs to 18870.
Am I understanding the problem correctly or do I need to tke a different approach?
As stated in the comments, the number of $5000 payments is two fewer than the total number of payments so the proper answer is:
$$ 10000P(X = 1) + \sum_{k = 2}^\infty \left[17500 + 5000(k - 2)\right]P(X = k) \approx 18,871.33 $$
[Closing, but made community wiki so as not to take credit for another's answer]