Equal Sum Partition in Odd Series

Let $k$ be the $m$-th term of the sequence of consecutive odd numbers, so $k = 2m - 1$. The sum of elements less than $k$ is the sum of the first $m-1$ odd numbers, which is $(m-1)^2$. The sum of all odd numbers from $1$ to $57$ (which has $29$ terms) is $29^2 = 841$. We are given: $\text{Sum}(< k) = \text{Sum}(> k) = S$ $2S + k = 841 \implies 2(m-1)^2 + (2m - 1) = 841$ $2(m^2 - 2m + 1) + 2m - 1 = 841 \implies 2m^2 - 2m + 1 = 841 \implies 2m^2 - 2m - 840 = 0$ $m^2 - m - 420 = 0 \implies (m - 21)(m + 20) = 0 \implies m = 21$ Thus, $k = 2(21) - 1 = 41$.