Question

If $\alpha$ not equal $1$ is any nth root of unity then $S=1+3\alpha +5{\alpha }^2+...$to $n$ terms equals

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

Answer 1

The correct option is B

$-\frac{2n}{(1-\alpha )}$

Explanation for the correct option:

Step 1. Find the value of $S=1+3\alpha +5{\alpha }^2+...$:

Let $S=1+3\alpha +5{\alpha }^2+....+\left(2n-1\right){\alpha }^{n-1}$ …..(1)

Multiply both side by $\alpha$, we get

$\Rightarrow$$S\alpha =\alpha +3{\alpha }^2+5{\alpha }^3+....+\left(2n-1\right){\alpha }^{n-1}+\left(2n-1\right){\alpha }^n$ …...(2)

Step 2. Subtract equation (2) from equation (1), we get

$\begin{array}{rcl}S(1-\alpha )& =& 1+2\alpha +2{\alpha }^2+...+2{\alpha }^{n-1}-\left(2n-1\right){\alpha }^n\\ & =& 1+2\alpha \left(1+\alpha +...+{\alpha }^{n-2}\right)-\left(2n-1\right){\alpha }^n\\ & =& 1+2\alpha \left(\frac{1-{\alpha }^{n-1}}{1-\alpha }\right)-\left(2n-1\right){\alpha }^n\\ & =& 1+\frac{2}{1-\alpha }\left(\alpha -{\alpha }^n\right)-\left(2n-1\right){\alpha }^n\\ & =& 1+\frac{2}{1-\alpha }\left(\alpha -1\right)-\left(2n-1\right){\alpha }^n\\ & =& -\left(2n-1\right)-1\\ & =& -2n\end{array}$

$\mathbf{\therefore }\mathit{S}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{-}\mathbf{2}\mathbf{n}}{\left(1-\alpha \right)}$

Hence, Option ‘B’ is Correct.