Question

If $\alpha$ is the n${}^{th}$ root of unity, then $1+2\alpha +3{\alpha }^2+....$ to $n$ terms equal to

• A. $\frac{-n}{(1-\alpha {)}^2}$
• B. $\frac{-n}{1-\alpha }$
• C. $\frac{-2n}{1-\alpha }$
• D. $\frac{-2n}{(1-\alpha {)}^2}$

Answer 1

The correct option is B $\frac{-n}{1-\alpha }$

$S=1+2\alpha +3{\alpha }^2+...n{\alpha }^{n-1}$
$\alpha S=\alpha +2{\alpha }^2+3{\alpha }^3+...(n-1){\alpha }^{n-1}+n{\alpha }^n$
$S(1-\alpha )=1+\alpha +{\alpha }^2+{\alpha }^3+...{\alpha }^{n-1}-n{\alpha }^n$
$S(1-\alpha )=\frac{1-{\alpha }^n}{1-\alpha }-n{\alpha }^n$
Now ${\alpha }^n=1$ since it is the $n^{th}$ root of unity.
Therefore,
$S(1-\alpha )=-n$
$S=\frac{-n}{1-\alpha }$