Question

If $\alpha$ be the $n^{th}$ root of unity then the sum of the series
$1+2\alpha +3{\alpha }^2+...n{\alpha }^{n-1}$ equals.

• A. $\frac{-n}{1-\alpha }$
• B. ${\frac{-n}{1-\alpha }}^2$
• C. $\frac{n}{1-\alpha }$
• D. None of these

Answer 1

Answer: (A)

$\frac{-n}{1-\alpha }$

$S=1+2\alpha +3{\alpha }^2+...n{\alpha }^{n-1}$
Here $S$ forms an A.G.P
Therefore
$S=1+2\alpha +3{\alpha }^2+...n{\alpha }^{n-1}$
$S\left(\alpha \right)=\alpha +2{\alpha }^2+3{\alpha }^3+...(n-1){\alpha }^{n-1}+n{\alpha }^n$
Hence
$S(1-\alpha )=1+\alpha +{\alpha }^2+....{\alpha }^{n-1}-n{\alpha }^n$
$S(1-\alpha )=\frac{1-{\alpha }^n}{1-\alpha }-n{\alpha }^n$
$S(1-\alpha )=0-n$
$S=\frac{-n}{1-\alpha }$
Hence, option 'A' is correct.