Question

If $\alpha =cos\frac{2\pi }{n}+isin\frac{2\pi }{n},$ then $\sum _{r=1}^n(ar+b){\alpha }^{r-1}$

• A. $\frac{na}{\alpha -1}$
• B. $\frac{nb}{\alpha -1}$
• C. $\frac{n\alpha }{\alpha -1}$
• D. $\frac{a}{\alpha -1}$

Answer 1

The correct option is A $\frac{na}{\alpha -1}$

$\alpha =cos\frac{2\pi }{n}+isin\frac{2\pi }{n}$
$\Rightarrow {\alpha }^n=cos2\pi +i\sin 2\pi$
$\Rightarrow {\alpha }^n=1$
Now, $\sum _{r=1}^n(ar+b){\alpha }^{r-1}$
$=(a+b)+(2a+b)\alpha +(3a+b){\alpha }^2+....+(na+b){\alpha }^{n-1}$
$=(a+2a\alpha +3a{\alpha }^2+....+na{\alpha }^{n-1})+b(1+\alpha +{\alpha }^2+....+{\alpha }^{n-1})$
$=(a+2a\alpha +3a{\alpha }^2+....+na{\alpha }^{n-1})(\because \alpha$ is $n^{th}$ root of unity.So $1+\alpha +{\alpha }^2+....{\alpha }^{n-1}=0)$
which is AGP.
Let $S=(a+2a\alpha +3a{\alpha }^2+....+na{\alpha }^{n-1})$
$\alpha S=a\alpha +2a{\alpha }^2+...+(n-1)a{\alpha }^{n-1}+na{\alpha }^n$
$\Rightarrow (1-\alpha )S=a+a\alpha +a{\alpha }^2+....a{\alpha }^{n-1}-na{\alpha }^n$
$\Rightarrow (1-\alpha )S=-na{\alpha }^n(\because \alpha$ is $n^{th}$ root of unity.So $1+\alpha +{\alpha }^2+....{\alpha }^{n-1}=0)$
$\Rightarrow S=\frac{na}{\alpha -1}$