Finding the total angular momentum for a system of particles The total angular momentum of the system can be found by forming the cross prodcut of ${\bf r}_i×{\bf p}_i$ and summing over i

Also since (N is toruqe):

In my textbook its says that last term in the 2nd equation can be expanded so

However i am having trouble understanding how the expanding the sum can give this?

Is it correct to assume this: $\sum_{i,j}a_i\times b_j= a_i\times b_j + a_j \times b_i $ (which seems wrong to me)

This is the whole extract from the textbook:

 A: It's just the third law ${\bf F}_{ij}= -{\bf F}_{ji}$, so
$$
\sum_{i,j} {\bf r}_i\times {\bf F}_{ij}= \frac 12  \sum_{i,j} ({\bf r}_i\times {\bf F}_{ij}+{\bf r}_j\times {\bf F}_{ji})  \quad (\hbox{just relabelling }i\leftrightarrow j)\\
=  \frac 12  \sum_{i,j} ({\bf r}_i\times {\bf F}_{ij}-{\bf r}_j\times {\bf F}_{ij})\\
\frac 12  \sum_{i,j} ({\bf r}_i-{\bf r}_j)\times {\bf F}_{ij}\\
= \sum_{\langle i,j\rangle} ({\bf r}_i-{\bf r}_j)\times {\bf F}_{ij},
$$
where $\langle i,j\rangle$ denotes the pair composed of $i$ and $j$. This means that   $\langle 1,2\rangle $ is not counted twice as being  $i=1,j=2$ and $i=2,j=1$, for example.
I bet your  book  goes on to say that $({\bf r}_i-{\bf r}_j)\times {\bf F}_{ij}=0$ because the force is parallel to the line separating the particles. If the book does do this, it is a common textbook cheat because there is no reason for this to be so if the forces arise from chemical bonds. The sum
$$
\sum_{i,j}({\bf r}_i-{\bf r}_j)\times {\bf F}_{ij}
$$
does vanish for a rigid body --- but the reason for this is not that  the individual $({\bf r}_i-{\bf r}_j)$  are  parallel to ${\bf F}_{ij}$.
This "parallel" argument is correct, however, if the forces are gravitational or electrostatic, as these are examples of the special case of central forces.
A: This eqution is correct,
$r_i \times F_{ji} + r_j \times F_{ij} = $
$r_i \times F_{ji} - r_j \times F_{ji} = $
$(r_i- r_j) \times  F_{ji} $
To get the second line we used Newton's third law i.e. $F_{ij} = - F_{ji}$.
Now let's look a the terms in the second sum,
$\sum_{i,j,i\neq j} r_i \times F_{ji} =$
$\frac{1}{2} \sum_{i,j,i\neq j} ( r_i \times F_{ji} + r_i \times F_{ji} ) =$
$\frac{1}{2} \Big( \sum_{i,j,i\neq j} r_i \times F_{ji}  +   \sum_{i,j,i\neq j} r_i \times F_{ji}  \Big) $
These two sums are independent of each other so you can relabel $i$ and $j$ as
follow
$\frac{1}{2} \Big( \sum_{i,j,i\neq j} r_i \times F_{ji}  +   \sum_{i,j,i\neq j} r_j \times F_{ij}  \Big) =$
$\frac{1}{2}\sum_{i,j,i\neq j}  \Big(  r_i \times F_{ji}  +  r_j \times F_{ij} \Big) =$
$\frac{1}{2}\sum_{i,j,i\neq j} (  r_i - r_j )\times F_{ji}$
