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Compute floor(1000 * T), where T = sum_{n=0}^{infinity} (n^2 + 1)/n!…
Question
Compute floor(1000 * T), where T = sum_{n=0}^{infinity} (n^2 + 1)/n! and floor denotes the greatest integer function. (Use 0! = 1.)
✓ Verified answer: 8154checked by our engine — not a guess
Step-by-step solution
Recall sum 1/n! = e, sum n/n! = e (shift index), and sum n^2/n! = 2e.
Indeed sum n^2/n! = sum n(n-1)/n! + sum n/n! = e + e = 2e.
So T = sum n^2/n! + sum 1/n! = 2e + e = 3e.
3e = 8.15484548..., so 1000*T = 8154.845..., and floor = 8154.
Final answer8154
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