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 $.

Thu, 30 Apr 2026 10:30:32 +0000

Content under review