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Suppose I have a system of ##N## mutually interacting particles, then the force on the ##i##-th particle due to the other ##N-1## particles is given by $$\mathbf{F}_{i}=\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}$$ If I then introduce a net external force, ##\mathbf{F}^{ext}## acting on the whole system, then the total force, ##\mathbf{F}## acting on the system is given by $$\mathbf{F}=\mathbf{F}^{ext}+\sum_{i=1}^{N}\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}$$ Concentrating on the double sum one can expand this as $$ \sum_{i=1}^{N}\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}=(\mathbf{F}_{12}+\mathbf{F}_{13}+\cdots)+(\mathbf{F}_{21}+\mathbf{F}_{23}+\cdots)+(\mathbf{F}_{31}+\mathbf{F}_{32}+\cdots)+(\mathbf{F}_{41}+\mathbf{F}_{42}+\cdots)+\cdots \\ =(\mathbf{F}_{12}+\mathbf{F}_{21}+\cdots)+(\mathbf{F}_{13}+\mathbf{F}_{31}+\cdots)+(\mathbf{F}_{14}+\mathbf{F}_{41}+\cdots)+\cdots \\ = \sum_{1<j}(\mathbf{F}_{1j}+\mathbf{F}_{j1})+\sum_{2<j}(\mathbf{F}_{2j}+\mathbf{F}_{j2})+\sum_{3<j}(\mathbf{F}_{3j}+\mathbf{F}_{j3})+\sum_{1<j}(\mathbf{F}_{4j}+\mathbf{F}_{j4})+ \cdots\\ =\sum_{i=1}^{N}\sum_{i<j}(\mathbf{F}_{ij}+\mathbf{F}_{ji})$$ From Newton's 3rd law we have that ##\mathbf{F}_{ij}=-\mathbf{F}_{ji}##, and so clearly this whole term vanishes (term-by-term). Hence we are left with the known result that the total force acting on a system of ##N## particles is equal to the external force acting on the system, thus enabling us to treat the system (as a whole) as a point particle.

Now, I'm sure there must be a more elegant why to arrive at this result than the way I have above, i.e. manipulating the double sum without having to expand in the way I did, but I can't seem to see the wood for the trees at the moment.

[One thought I had was to write $$\sum_{i=1}^{N}\sum_{j=1, i\neq j}^{N}\mathbf{F}_{ij}=\sum_{i=1}^{N}\left(\sum_{i<j}\mathbf{F}_{ij}+\sum_{i>j}\mathbf{F}_{ij}\right)$$ but I'm a little unsure about this. I mean, can one legitimately write ##\sum_{i>j}\mathbf{F}_{ij}=\sum_{i<j}\mathbf{F}_{ji}##?]

If anyone can provide a more elegant approach I'd much appreciate it.