Let us consider $\vec{F_i}$ be the resultant external force acting on the ith particle of a system of n particles and $\vec{F_{ij}}$ be the internal force on the ith particle due to the jth particle.

So the resultant internal force on the ith particle due to all other particle is $\displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}}$, where $i\neq{j}$.

If $\vec{p_i}$ be the linear momentum of the ith particle of mass $m_i$ moving with velocity $\vec{v_i}$ and $\vec{r_i}$ be the position vector of ith particle then the total force on the ith particle is given by,

$\vec{F_i}+\displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}}=\displaystyle{\frac{d}{dt}\vec{p_i}}=\frac{d}{dt}(m_i\vec{v_i})$

or, $\vec{F_i}+\displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}}=\frac{d^2}{dt^2}(m_i\vec{r_i})\tag{1}$

Summing over $i$ in equation (1) we get,

$\displaystyle{\sum_{i=1}^{n}}\vec{F_i}+\displaystyle{\sum_{i=1}^{n}}\displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}}=\frac{d^2}{dt^2}\left[\displaystyle{\sum_{i=1}^{n}}(m_i\vec{r_i})\right]$

Now according to Newton’s 3rd law, $\vec{F_{ij}}=-\vec{F_{ji}}$, so $\displaystyle{\sum_{i=1}^{n}}\displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}}=0$

Therefore,

$\displaystyle{\sum_{i=1}^{n}}\vec{F_i}=\frac{d^2}{dt^2}\left[\displaystyle{\sum_{i=1}^{n}}(m_i\vec{r_i})\right]$

or, $\vec{F}=\frac{d^2}{dt^2}(M\vec{R})$

where, $\displaystyle{\vec{R}=\frac{\displaystyle{\sum_{i=1}^{n}}m_i\vec{r_i}}{M}}$ is the position vector of centre of mass of the system and $\displaystyle{(M=\sum_{i=1}^{n}m_i)}$ is the total mass of the system.

or, $\displaystyle{\vec{F}=M\frac{d^2\vec{R}}{dt^2}}$

$\vec{F}$ is the total force acting on the system of particles.