Question

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

• A. $\frac{n(n+1)a}{2}$
• B. $\frac{nb}{1-n}$
• C. $\frac{na}{\omega -1}$
• D. $noneofthese$

Answer 1

Answer: (C)

$\frac{na}{\omega -1}$

Upon expanding, we get
$(a+b)+(2a+b)w+(3a+b)w^2+...(na+b)w^{n-1}$
$=a(1+2w+3w^2+...n{w}^{n-1})+b(1+w+w^2+...w^{n-1})$
$=a(1+2w+3w^2+...n{w}^{n-1})+b\left(\frac{1-w^n}{1-w}\right)$
$=a(1+2w+3w^2+...n{w}^{n-1})+0$

Let
$S=a(1+2w+3w^2+...n{w}^{n-1})$
$Sw=a(w+2w^2+3w^3+...(n-1)w^{n-1}-n{w}^n)$
$S(1-w)=a(1+w+w^2....w^{n-1})-an{w}^n$
$S(1-w)=a\left(\frac{1-w^n}{1-w}\right)-an{w}^n$
$S(1-w)=a\left(0\right)-an$
$S=-\frac{an}{1-w}=\frac{an}{w-1}$
Hence, option 'C' is correct.