Question

If $\alpha$ is the $n^{th}$ root of unity, then $1+2\alpha +3{\alpha }^2+...$ 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 $-\frac{n}{(1-\alpha )}$

$S=1+2\alpha +3{\alpha }^2+....n{\alpha }^{n-1}$
$S\alpha =\alpha +2{\alpha }^2+3{\alpha }^3...(n-1){\alpha }^{n-1}+n{\alpha }^n$
$S(1-\alpha )=1+\alpha +{\alpha }^2+...{\alpha }^{n-1}-n{\alpha }^n$
$S(1-\alpha )=\frac{{\alpha }^n-1}{\alpha -1}-n{\alpha }^n$
Since $\alpha$ is the nth root of unity, hence ${\alpha }^n=1$
Thus
$S(1-\alpha )=\frac{{\alpha }^n-1}{\alpha -1}-n{\alpha }^n$
$S(1-\alpha )=\frac{1-1}{\alpha -1}-n$
$S(1-\alpha )=-n$
$S=-\frac{n}{1-\alpha }$