Sum the odd square numbers less than N
Jelly, 6 bytes

½Ċ|1c3


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Background

For all positive integers k, we have 1² + 3² + ⋯ + (2k - 1)² = k(2k - 1)(2k +1) ÷ 3.

Since there are m C r = m! ÷ ((m-r)!r!) r-combinations of a set of m elements, the above can be calculated as (2k + 1) C 3 = (2k + 1)2k(2k - 1) ÷ 6 = k(2k - 1)(2k + 1) ÷ 3.

To apply the formula, we must find the highest 2k + 1 such that (2k - 1)² &lt; n. Ignoring the parity for a moment, we can compute the highest m such that (m - 1)² &lt; n as m = ceil(srqt(n)). To conditionally increment m if it is even, simply compute m | 1 (bitwise OR with 1).

How it works

½Ċ|1c3  Main link. Argument: n

½       Compute the square root of n.
 Ċ      Round it up to the nearest integer.
  |1    Bitwise OR with 1 to get an odd number.
    c3  Compute (2k + 1) C 3 (combinations).