Square Numbers which are Sum of Sequence of Odd Cubes/Mistake

Source Work

 * The Dictionary
 * $1225$
 * $1225$

Mistake

 * $35^2 = 1^3 + 3^3 + 5^3 + 7^3 + 9^3$. The next such sum is $1^3 + \ldots 29^3$.

Correction
This is wrong, as can be seen by the fact that:
 * $1^3 + \cdots + 29^3 = 101 \, 025$

which is not a square number.

In fact, the next such sum is $1^3 + 3^3 + 5^3 + \dotsb + 55^3 + 57^3$.

That is:
 * $1^3 + \paren {2 \times 2 - 1}^3 + \paren {2 \times 3 - 1}^3 + \dotsb + \paren {2 \times 28 - 1}^3 + \paren {2 \times 29 - 1}^3$

What seems to have done (and is seems to be a common confusion) is confuse:
 * $\ds \sum_{j \mathop = 1}^{29} \paren {2 j - 1}^3$

with:
 * $1^3 + 3^3 + \dotsb + 27^3 + 29^3$

which is not the same thing at all.