If $n \leq 3$ and $a, a_{1}, a_{2}, . . ., a_{n - 1}$ are $n$ th roots of unity, then find the sum $\sum 1 \leq i < j \leq n - 1 α_{i} α_{j}$ .

Open in App

Solution

We have,
$z^{n} - 1 = (z - 1) (z - a_{1}) (z - a_{2}) . . . (z - a_{n} - 1)$
Now the sum of products taken two at a time,
$S = \sum 1 \leq i < j \leq n - 1 α_{i} α_{j} + α_{1} + α_{2} + . . . + α_{n - 1}$ .
Now $S = 0$ as coefficient of $z^{n - 1}$ is zero and
$a_{1} + a_{2} + . . . + a_{n - 1} = - 1$
$∴ 0 = \sum 1 \leq i < j \leq n - 1 α_{i} α_{j} - 1$
$\Rightarrow \sum_{1 \leq i < j \leq n - 1} α_{i} α_{j} = 1$ .
Ans: 1

Suggest Corrections

Similar questions

Q. If

a_{1}, a_{2}, . . . . a_{n - 1}

are

n

th roots of unity then prove that
(i)

(1 - a_{1}) (1 - a_{2}) (1 - a_{n - 1}) . . . = n

(ii)

1 + a_{1} + a_{2} + . . . . + a_{n - 1} = 0

(iii)

\frac{1}{2 - a_{1}} + \frac{1}{2 - a_{2}} + . . . . \frac{1}{2 - a_{n - 1}} = \frac{(n - 2) 2^{n - 1} + 1}{2^{n} - 1}

Q. If

1, a_{1}, a_{2}, . a_{n - 1}

, are

n

roots of unity, then the value of

(1 - a_{1}) (1 - a_{2}) (1 - a_{3}) . (1 - a_{n - 1})

is equal to ?

α_{0}, α_{1}, α_{2}, . . . . . . . α_{n - 1}

be the

n, n^{t h}

roots of the unity, then the value

n - 1 \sum i = 0 \frac{α_{i}}{3 - α_{i}}

is equal to

If $a_{0}, a_{1}, a_{2}, . . . . . . . . . . . ., a_{n - 1}$ are nth roots of unity, then $a_{k}$ = cos $\frac{2 k π}{n}$ + i sin $\frac{2 k π}{n}$ where $0 \leq k \leq n - 1$ .

Also then Value of $2^{n - 1}$ sin $(\frac{π}{n})$ sin $(\frac{2 π}{n})$ .........sin $(\frac{n - 1}{n} π)$ is

Q. If

1, w, w^{2}, . . . w^{n - 1}

are the

n^{t h}

roots of unity, then

(1 - w) (1 - w^{2}) . . . (1 - w^{n - 1}) i s

Join BYJU'S Learning Program

If n≤3 and a,a1,a2,...,an−1 are nth roots of unity, then find the sum ∑1≤i<j≤n−1αiαj.

We have,zn−1=(z−1)(z−a1)(z−a2)...(z−an−1)Now the sum of products taken two at a time,S=∑1≤i<j≤n−1αiαj+α1+α2+...+αn−1.Now S=0 as coefficient of zn−1 is zero anda1+a2+...+an−1=−1∴0=∑1≤i<j≤n−1αiαj−1⇒∑1≤i<j≤n−1αiαj=1.Ans: 1

If $n \leq 3$ and $a, a_{1}, a_{2}, . . ., a_{n - 1}$ are $n$ th roots of unity, then find the sum $\sum 1 \leq i < j \leq n - 1 α_{i} α_{j}$ .