JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Medium

1 of 1

Analyzing +4 | -1

\mathrm{\alpha } \ne 1

is any

nth

root of unity, then

S = 1 + 3\alpha + 5{\alpha ^2} \ldots

upon

n

terms, is equal to

View answer and explanation

Correct option: B

- \frac{{2n}}{{1 - \alpha }}

Option B (

- \frac{{2n}}{{1 - \alpha }}

) is correct because it is derived by multiplying the original series by alpha, subtracting, and using the properties of the nth root of unity.

More Questions from COMPLEX NUMBERS

Practice All Chapter MCQs →

{\left| {1 - i} \right|^x} = {2^x}

|z|=1

z \neq \pm 1

z, \frac{z}{1-z^2}

\left|z_r-r\right| \leq r, \forall r=1,2,3, \ldots, n

\left|\sum\limits_{r=1}^n z_r\right|

(z+a b)^3=a^3

a \neq 0

\alpha, \beta

\gamma

|\alpha-\beta|

{z^2} + {\left| z \right|^2} = 0,

z \in C

2x = - 1 + \sqrt 3 i

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}