The given expression is ${\left(a+b+c\right)}^n+{\left(a+b-c\right)}^n$

Suppose that $a+b=x$

Hence the above expression becomes $(\mathrm{x}+\mathrm{c}{)}^{\mathrm{n}}+(\mathrm{x}-\mathrm{c}{)}^{\mathrm{n}}$, this expression has $\left(\frac{n}{2}+1\right)$ terms

The expansion is $2\left[(\mathrm{a}+\mathrm{b}{)}^{\mathrm{n}}{+}^{\mathrm{n}}{\mathrm{C}}_2(\mathrm{a}+\mathrm{b}{)}^{\mathrm{n}-2}{\mathrm{c}}^2+\dots \dots ..+{\mathrm{c}}^{\mathrm{n}}\right]$

Hence, the number of terms is $2\left[\right(\mathrm{n}+1)\text{ trems }+(\mathrm{n}-1)\text{ terms }+\dots +1]$

This is in arithametic progression, having first term 1, and common difference is 2, number of terms is $\left(\frac{n}{2}+1\right)$

Hence the sum of first $\left(\frac{n}{2}+1\right)$ terms is 

                                $\begin{array}{rcl}\frac{\left(\frac{n}{2}+1\right)}{2}\left[2+\left(\frac{n}{2}\right)2\right]& =& \frac{n+2}{4}\left(n+2\right)\\ & =& \frac{{\left(n+2\right)}^2}{4}\\ & =& \boxed{{\left(\frac{n+2}{2}\right)}^2}\end{array}$