Summation Equation Comparison: Spivak's Calculus Answer Book vs Speculation

[julypraise]

Homework Statement

The Spivak's Calculus Answer Book (3ed) states that, on page 17,

\sum_{i \neq j} (x_{i}^{2}y_{j}^2 - x_{i}y_{i}x_{j}y_{j}) = 2\sum_{i < j}(x_{i}^{2}y_{j}^2 + x_{j}^{2}y_{i}^2 - x_{i}y_{i}x_{j}y_{j})

But as I speculate, I've got the following:

\sum_{i \neq j} (x_{i}^{2}y_{j}^2 - x_{i}y_{i}x_{j}y_{j}) = \sum_{i < j}(x_{i}^{2}y_{j}^2 + x_{j}^{2}y_{i}^2 - 2x_{i}y_{i}x_{j}y_{j})

Could you check which is right?

Thanks.

Homework Equations

The Attempt at a Solution

\sum_{i \neq j} (x_{i}^{2}y_{j}^2 - x_{i}y_{i}x_{j}y_{j}) = \sum_{i < j}(x_{i}^{2}y_{j}^2 + x_{j}^{2}y_{i}^2 - 2x_{i}y_{i}x_{j}y_{j})

 

[CompuChip]

I suspect that the answer in the book is wrong.
You can easily write it out manually for i, j running through {1, 2}.

It's also possible to prove using
\sum_{i \neq j} a_{ij} = \sum_{i < j} a_{ij} + \sum_{i > j} a_{ij}
where
\sum_{i > j} a_{ij} = \sum_{j > i} a_{ji} = \sum_{i < j} a_{ji}