Sum Odd Squares 64Eda5

1. **State the problem:** We want to find the sum of all odd square numbers among the first 870,000 square numbers. 2. **Understanding odd squares:** Odd squares come from squaring odd integers. The $n$-th odd number is $2n - 1$, so the $n$-th odd square is $(2n - 1)^2$. 3. **Count of odd squares in first 870,000 squares:** The first 870,000 square numbers are $1^2, 2^2, 3^2, ..., 870000^2$. Among these, the odd squares correspond to odd integers from 1 to 870,000. The number of odd integers up to 870,000 is: $$\text{count} = \frac{870000 + 1}{2} = 435000.5$$ Since count must be an integer, we take the floor: $$\text{count} = 435000$$ 4. **Sum of the first $m$ odd squares:** We want to find: $$S = \sum_{n=1}^{435000} (2n - 1)^2$$ Expand the square: $$S = \sum_{n=1}^{435000} (4n^2 - 4n + 1) = 4\sum_{n=1}^{435000} n^2 - 4\sum_{n=1}^{435000} n + \sum_{n=1}^{435000} 1$$ 5. **Use formulas for sums:** - Sum of first $m$ integers: $$\sum_{n=1}^m n = \frac{m(m+1)}{2}$$ - Sum of first $m$ squares: $$\sum_{n=1}^m n^2 = \frac{m(m+1)(2m+1)}{6}$$ - Sum of 1 repeated $m$ times: $$\sum_{n=1}^m 1 = m$$ 6. **Calculate each sum for $m=435000$:** $$\sum_{n=1}^{435000} n = \frac{435000 \times 435001}{2}$$ $$\sum_{n=1}^{435000} n^2 = \frac{435000 \times 435001 \times 870001}{6}$$ $$\sum_{n=1}^{435000} 1 = 435000$$ 7. **Substitute back:** $$S = 4 \times \frac{435000 \times 435001 \times 870001}{6} - 4 \times \frac{435000 \times 435001}{2} + 435000$$ 8. **Simplify terms:** $$S = \frac{4 \times 435000 \times 435001 \times 870001}{6} - 2 \times 435000 \times 435001 + 435000$$ 9. **Final answer:** The sum of all odd squares among the first 870,000 square numbers is: $$\boxed{S = \frac{4 \times 435000 \times 435001 \times 870001}{6} - 2 \times 435000 \times 435001 + 435000}$$ This formula can be computed with a calculator or software for the exact numeric value.