newtonian mechanics - The sum of all external forces acting on all the particles is equal to the total external force applied to the system of particles. Why?

As per the book:

$\sum_{j=1}^N f_{j}^{ext}$+$\sum_{j=1}^N f_{j}^{int}$=$\sum_{j=1}^N dp_{j}/dt$

The first term, $\sum_{j=1}^N f_{j}^{int}$, is the sum of all internal forces acting on all the particles.

$\sum_{j=1}^N f_{j}^{int}$=$0$ (because according to newton's 3rd law, every particle will have the reaction force as a counteracting force within the system-which I understood)

The second term, $\sum_{j=1}^N f_{j}^{ext}$, is the sum of all external forces acting on all the particles. It is the total external force $F_{ext}$ acting on the system.:

$\sum_{j=1}^N f_{j}^{ext}$ ≡$ F_{ext}$

However, I didn't really get how can we say that the second term is equal to total external force. It would be appreciated if it could be explained with the help of some graph or figure by choosing small number of particle and showing that the sum of force of interaction of these particle is indeed the total external force applied to the system.

Answer

Since you seem happy about the internal forces let's ignore them and just set them all to zero so only consider external forces.

The total momentum $P$ is just the sum of all the individual momenta:

$$ P = \sum p_i $$

and we can differentiate both sides of this to get:

$$ \frac{dP}{dt} = \sum \frac{dp_i}{dt} $$

For any object, simple or composite, force is the rate of change of momentum - that is just Newton's second law. Now, the left side of the equation above is the rate of change of total momentum so that's the total force:

$$ \frac{dP}{dt} = F_\text{tot} $$

The right side is the rate of change of momenta of the individual particles so that's the force on the individual particles:

$$ \frac{dp_i}{dt} = f_i $$

So we end up with:

$$ F_\text{tot} = \sum f_i $$