Question

If $\omega$ is a complex nth root of unity, then ${\sum }_{r=1}^n(ar+b){\omega }^{r-1}$ is equal to

• A. $\frac{n(n+1)a}{2}$
• B. $\frac{ab}{1-n}$
• C. $\frac{na}{\omega -1}$
• D. None of these

Answer 1

The correct option is C $\frac{na}{\omega -1}$

Expanding, we get
$(a+b)+(2a+b)w^1+(3a+b)w^2+....(na+b)w^{n-1}$
$=a(1+2w+3w^2+...n{w}^{n-1})+b(1+w+w^2+...w^{n-1})$
Let $S=1+2w+3w^2+...n{w}^{n-1}$
$Sw=w+2w^2+3w^3+...(n-1)w^{n-1}+n{w}^n$
$S(1-w)=1+w+w^2+....w^{n-1}-n{w}^n$
$=\frac{1-w^n}{1-w}-\frac{n{w}^n}{1-w}$
Hence
$a(1+2w+3w^2+...n{w}^{n-1})+b(1+w+w^2+...w^{n-1})$
$=a(\frac{1-w^n}{1-w}-\frac{n{w}^n}{1-w})+b\frac{1-w^n}{1-w}$
$=a(0-\frac{n}{1-w})+0$
$=\frac{an}{w-1}$