Question

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

Let $S=1+2\alpha +3{\alpha }^2+\dots +n{\alpha }^{n-1}$
or, $\alpha S=\alpha +2{\alpha }^2+3{\alpha }^3+\dots +(n-1){\alpha }^{n-1}+n{\alpha }^n$
On substracting, we get
$S(1-\alpha )=1+[\alpha +{\alpha }^2+\dots +{\alpha }^{n-1}]-n{\alpha }^n$
$=1+\frac{\alpha (1-{\alpha }^{n-1})}{1-\alpha }-n{\alpha }^n$
or, $S=\frac{1}{1-\alpha }+\frac{\alpha -{\alpha }^n}{(1-\alpha {)}^2}-\frac{n{\alpha }^n}{1-\alpha }$
$=\frac{1}{1-\alpha }+\frac{\alpha -1}{(1-\alpha {)}^2}-\frac{n}{1-\alpha }$
$=-\frac{n}{1-\alpha }$