If - is an imaginary cube root of unity- then the value of 2-2- 2-23-3- 2-n-1n-n- 2 is

The correct option is A n^24(n+1)^2 n general term of given sequence is #160; T_r=(r 1)(r ω)(r ω^2)=(r 1)[r^2 r(ω+ω^2)+ω^3]=(r 1)(r^2+r+1)=r^3 1 Si ...

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Standard XII

Mathematics

Properties of Iota

If ω is an ...

Question

If $ω$ is an imaginary cube root of unity, then the value of
$(2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + . . . . . + (n - 1) (n - ω) (n - ω^{2})$ is

\frac{n^{2}}{4} (n + 1)^{2} - n

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\frac{n^{2}}{4} (n + 1)^{2} + n

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\frac{n^{2}}{4} (n + 1)^{2}

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\frac{n^{2}}{4} (n + 1) - n

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Solution

The correct option is A $\frac{n^{2}}{4} (n + 1)^{2} - n$
general term of given sequence is
$T_{r} = (r - 1) (r - ω) (r - ω^{2}) = (r - 1) [r^{2} - r (ω + ω^{2}) + ω^{3}] = (r - 1) (r^{2} + r + 1) = r^{3} - 1$
Since $1 + ω + ω^{2} = 1, ω^{3} = 1, a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$
Hence required sum $= n \sum r = 1 T_{r} = n \sum r = 1 r^{3} - n \sum r = 1 1 = \frac{n^{2}}{4} (n + 1)^{2} - n$ , (use special sum of series formula)

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Similar questions

Q. If

ω

is an imaginary cube root of unity, then the value of

(2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + . . . + (n - 1) (n - ω) (n - ω^{2})

Q. If

ω

is an imaginary cube root of unity, then the value of the expression

1 (2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + . . . . + (n - 1) (n - ω) (n - ω^{2})

Q. If

ω

is an imaginary cube root of

1

, then the value of

1 (2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + . . . . + (n - 1) (n - ω) (n - ω^{2})

Q. If

ω

is an imaginary cube root of unity then expression

1. (2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + . . . . . + (n - 1) (n - ω) (n - ω^{2})

is equal to

Q. If

ω

is a complex cube root of unity, then the value of the expression

1 (2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + . . . + (n - 1) (n - ω) (n - ω^{2})

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If ω is an imaginary cube root of unity, then the value of(2−ω)(2−ω2)+2(3−ω)(3−ω2)+.....+(n−1)(n−ω)(n−ω2) is

The correct option is A n24(n+1)2−ngeneral term of given sequence is Tr=(r−1)(r−ω)(r−ω2)=(r−1)[r2−r(ω+ω2)+ω3]=(r−1)(r2+r+1)=r3−1Since 1+ω+ω2=1,ω3=1,a3−b3=(a−b)(a2+ab+b2)Hence required sum =n∑r=1Tr=n∑r=1r3−n∑r=11=n24(n+1)2−n, (use special sum of series formula)

If $ω$ is an imaginary cube root of unity, then the value of
$(2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + . . . . . + (n - 1) (n - ω) (n - ω^{2})$ is