Write the expression for the centre of mass of a system of $n$ particles and derive the formula for the force acting on its centre of mass.

The position vector of the centre of mass of a system of $n$ particles is given by:
$\vec{R}_{cm} = \frac{m_{1} \vec{r}_{1} + m_{2} \vec{r}_{2} + \dots + m_{n} \vec{r}_{n}}{m_{1} + m_{2} + \dots + m_{n}} = \frac{1}{M} \sum_{i=1}^{n} m_{i} \vec{r}_{i}$
where $M = \sum m_{i}$ is the total mass of the system.
To find the force,we differentiate the position vector with respect to time to get velocity:
$\vec{V}_{cm} = \frac{d\vec{R}_{cm}}{dt} = \frac{1}{M} \sum_{i=1}^{n} m_{i} \vec{v}_{i}$
$M \vec{V}_{cm} = \sum_{i=1}^{n} m_{i} \vec{v}_{i} \quad (1)$
Differentiating equation $(1)$ with respect to time $t$ gives the acceleration:
$M \frac{d\vec{V}_{cm}}{dt} = \sum_{i=1}^{n} m_{i} \frac{d\vec{v}_{i}}{dt}$
$M \vec{A}_{cm} = \sum_{i=1}^{n} m_{i} \vec{a}_{i}$
Since the force on the $i$-th particle is $\vec{F}_{i} = m_{i} \vec{a}_{i}$,the total external force $\vec{F}_{ext}$ acting on the system is the sum of all individual forces:
$\vec{F}_{ext} = \sum_{i=1}^{n} \vec{F}_{i} = \sum_{i=1}^{n} m_{i} \vec{a}_{i}$
Therefore,$M \vec{A}_{cm} = \vec{F}_{ext}$.
This shows that the total external force on the system is equal to the product of the total mass of the system and the acceleration of its centre of mass.