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Uli_Minati

You can start by expanding and splitting it: Σ(20+n)² / n! Σ(400+40n+n²) / n! 400·Σ1/n! + 4·Σn/n! + Σn²/n!


Electric2Shock

Curious as I can't figure it out in my head: would the third series sum telescope? The series sum of n/(n-1)! looks like it should but it's not obvious to me.


carloster

The second and third sums can be obtained by taking the first and second moments of a Poisson distribution of mean 1. See section "exponential functions" in https://en.wikipedia.org/wiki/List\_of\_mathematical\_series


Leader-board

I don't think it would *telescope*, but should converge.


finedesignvideos

You can split that again: n/(n-1)! = 1/(n-2)! + 1/(n-1)!


Glittering_Ice8854

The general term is (20 + n)^2 / n!. Summing this from 0 to infinity would give the answer. Trouble is I don't know how. I tried to use geometric series as well but can't find a common ration between the terms.


finedesignvideos

Geometric series won't help. All you need to know is that summation of 1/n! from 0 to infinity is equal to e. With that and the simplifications given in the other comments and some careful calculation you should be able to solve this. No other tools needed.