Solution: We compute $ \sum_{n=1}^{10} P(n) = \sum_{n=1}^{10} (n^2 + 3n + 5) = \sum_{n=1}^{10} n^2 + 3\sum_{n=1}^{10} n + \sum_{n=1}^{10} 5 $.

Sat, 02 May 2026 20:46:18 +0000

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