Question

$(1-\omega +{\omega }^2)(1-{\omega }^2+{\omega }^4)(1-{\omega }^4+{\omega }^8)......$to 2n factors =

• A. 2
• B. $2^{2n}$
• C. 2n
• D. none of these

Answer 1

The correct option is B $2^{2n}$

Given,

$(1-\omega +{\omega }^2)(1-{\omega }^2+{\omega }^4)(1-{\omega }^4+{\omega }^8)$...... to $2n$

we have,

$1+w+w^2=0$

and $w^3=1$

$\Rightarrow (-w-w)(-w^2-w^2)(-w-w)....................2n$

$=(-2w)(-2w^2)(-2w).............2n$

here, we can see, 2 consecutive terms are same and are even, so we get,

$=\left(4w^3\right)\left(4w^3\right)\left(4w^3\right).........2n$

$=4\times 4\times .........2n$

$=2^2.........2n$

for $2n$ terms, we get,

$=2^{2n}$