SOLUTION

Let

$$S=\underset{r=1}{\overset{n}{\mathop \sum }}\,\left( ar+b \right){{\omega }^{r-1}}=a\underset{r=1}{\overset{n}{\mathop \sum }}\,r{{\omega }^{r-1}}+b\underset{r=1}{\overset{n}{\mathop \sum }}\,{{\omega }^{r-1}}$$

= a { 1 + 2ω + 3ω² + 4ω³ + ............. + nω^n-1 } + b { 1 + ω + ω² + ω³ + ............. + ω^n-1 } ............(i) 

As 1 + 2ω + 3ω² + ............. + nω^n-1 is an A.G.P.,

It is equal to $$\cfrac{1+\omega +{{\omega }^{2}}+\ldots \ldots \ldots \ldots .={{\omega }^{n-1}}}{1-\omega }-\cfrac{n{{\omega }^{n}}}{1-\omega }$$

Hence , $$S=a\left\{ \cfrac{1+\omega +{{\omega }^{2}}+\ldots \ldots \ldots \ldots .+{{\omega }^{n-1}}}{1-\omega }-\cfrac{n{{\omega }^{n}}}{1-\omega } \right\}$$

+ b { 1 + ω + ω² + ω³ + ............. + ω^n-1 }

$$=a\left\{ \cfrac{0}{1-\omega }-\cfrac{n}{1-\omega } \right\}+0=\cfrac{na}{\omega -1}$$

(∴ω is an imaginary n^th root of unity )
∴ ωⁿ = 1 and 1 + ω + ω² + ............. + ω^n-1 = 0