If n- 3 and 1-1- -2- - - -n-1 are nth roots of unity- then the value of -1 - i - j - n-1-i-j is

The correct option is A 1 Given that 1,α #160;_ 1 ,α #160;_ 2 ,....,α #160;_ n are n th roots of unity ⇒ x^n=1 So the sum of roots is ...

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Standard XII

Mathematics

Range

If n≥ 3 and...

Question

If $n \geq 3$ and $1, α_{1}, α_{2}, . . ., α_{n - 1}$ are $n t h$ roots of unity, then the value of $\sum 1 \leq i < j \leq n - 1 α_{i} α_{j}$ is

0

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1

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- 1

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(- 1)^{n}

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Solution

The correct option is A $1$
Given that $1, α_{1}, α_{2}, . . . ., α_{n}$ are $n$ th roots of unity
$\Rightarrow x^{n} = 1$
So the sum of roots is $0$
$\Rightarrow 1 + α_{1} + α_{2} + . . . ., {+ α}_{n} = 0$
Sum of product of roots taken two at a time is $0$
$\Rightarrow 1 (α_{1} + α_{2} + . . . ., {+ α}_{n}) + \sum_{1 \leq i < j \leq n - 1} α_{i} α_{j} = 0$
$\Rightarrow \sum_{1 \leq i < j \leq n - 1} α_{i} α_{j} = - (α_{1} + α_{2} + . . . ., {+ α}_{n}) = - (- 1) = 1$

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If n≥3 and 1,α1,α2,...,αn−1 are nth roots of unity, then the value of ∑1≤i<j≤n−1αiαj is

The correct option is A 1Given that 1,α1,α2,....,αn are nth roots of unity⇒xn=1So the sum of roots is 0⇒1+α1+α2+....,+αn=0Sum of product of roots taken two at a time is 0⇒1(α1+α2+....,+αn)+∑1≤i<j≤n−1αiαj=0⇒∑1≤i<j≤n−1αiαj=−(α1+α2+....,+αn)=−(−1)=1

If $n \geq 3$ and $1, α_{1}, α_{2}, . . ., α_{n - 1}$ are $n t h$ roots of unity, then the value of $\sum 1 \leq i < j \leq n - 1 α_{i} α_{j}$ is