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If ω is an imaginary cube root of unity then the value of cos[ $\frac{π}{225} \sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω}$ )] is: -

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-1

\frac{- 1}{2}

\frac{1}{2}

answer is D.

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Detailed Solution

Here, we have been provided with an imaginary cube root of unity ω and we have been asked to find the value of the expression: - cos[

\frac{π}{225} \sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω})

]
Now, we know that ω=

\frac{- 1 + \sqrt{3} i}{2}

and its conjugate (

\underset{̲}{ω}

) is given as: -
⇒

\underset{̲}{ω}

\frac{- 1 - \sqrt{3} i}{2}

ω^{2}

⇒

\underset{̲}{ω}

ω^{2}

- (1)
Now, let us find the value of the expression: -

\sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω})

. So, using equation (1), we get,
⇒

\sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω}) = \sum_{r = 1}^{10} (r - ω) (r - ω^{2})

⇒

\sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω}) = \sum_{r = 1}^{10} (r^{2} - rω - r ω^{2} + ω^{3})

⇒

\sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω}) = \sum_{r = 1}^{10} (r^{2} - r (ω + ω^{2}) + ω^{3})

Now, applying the formulas: - 1+ω+

ω^{2}

=0 and

ω^{3} = 1

, we get,
⇒

\sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω}) = \sum_{r = 1}^{10} (r^{2} - r (- 1) + 1)

⇒

\sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω}) = \sum_{r = 1}^{10} (r^{2} + r + 1)

Breaking the terms with summation sign, we get,
⇒

\sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω}) = \sum_{r = 1}^{10} r^{2} + \sum_{r = 1}^{10} r + \sum_{r = 1}^{10} 1

Here, we can clearly, see that the first term in R.H.S., i.e.

\sum_{r = 1}^{10} r^{2}

is the summation of squares of first 10 natural numbers whose sum is given as

\frac{n (n + 1) (2 n + 1)}{6}

, where n = 10. The second term is the summation of first 10 natural numbers, i.e.

\sum_{r = 1}^{10} r

, whose sum is given as

\frac{n (n + 1)}{2}

, where n = 10. Now, the third term is the summation of a constant k = 1, i.e.

\sum_{r = 1}^{10} 1

, whose sum is given as n, where n = 10. So, simplifying the summation, we get,
⇒

\sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω}) = \frac{10 \times (10 + 1) \times (20 + 1)}{6} + \frac{10 \times (10 + 1)}{2} + 10

⇒

\sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω}) = 450

Multiplying

\frac{π}{225}

both sides, we get,
⇒

\frac{π}{225} \sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω})

\frac{π}{225}

×450
⇒

\frac{π}{225} \sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω})

=2π
Now, taking cosine function both sides, we get,
⇒cos[

\frac{π}{225} \sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω})] =

cos2π
⇒ cos[

\frac{π}{225} \sum_{r = 1}^{10} (r - ω) (r - \underset{̲}{ω})

] = 1
Hence, option (4) is our Answer

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