Solution: First, compute the sum $ \sum_{k=1}^{6} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91 $. Next, compute the product $ \prod_{j=1}^{3} (2j + 1) = (2\cdot1 + 1)(2\cdot2 + 1)(2\cdot3 + 1) = 3 \cdot 5 \cdot 7 = 105 $. Now multiply the results: $ 91 \times 105 = 9105 $. Therefore, the value is $ \boxed{9105} $.

Mon, 04 May 2026 19:08:07 +0000

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