If xn - 1 - x - 1x - ax - a2-x - an - 1- then prove that -r - 1n - 11 2 - ar - n2n - 12n - 1 - 1-

x^n 1=(x 1)(x a)(x a^2)⋯(x a^n 1) applying logarithm on both sides log (x^n 1)=log (x 1)+log(x a)+log(x a^2)+⋯+log(x a^n 1) Differentiating on both sid ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Properties of Cube Root of a Complex Number

If xn - 1 =...

Question

If $x^{n} - 1 = (x - 1) (x - a) (x - a^{2}) . . . . . . . . (x - a^{n - 1})$ , then prove that $n - 1 \sum r = 1 \frac{1}{2 - a_{r}} = \frac{n 2^{n - 1}}{2^{n} - 1} - 1$ .

Open in App

Solution

$x^{n} - 1 = (x - 1) (x - a) (x - a^{2}) \dots (x - a^{n - 1})$
applying logarithm on both sides
$log (x^{n} - 1) = log (x - 1) + log (x - a) + log (x - a^{2}) + \dots + log (x - a^{n - 1})$
Differentiating on both sides
$\frac{n x^{n - 1}}{x^{n} - 1} = \frac{1}{x - 1} + \frac{1}{x - a} + \frac{1}{x - a^{2}} + \dots + \frac{1}{x - a^{n - 1}}$
Put $x = 2$
$⟹ 1 + n - 1 \sum r = 1 \frac{1}{2 - a^{r}} = \frac{n {.2}^{n - 1}}{2^{n} - 1} ⟹ n - 1 \sum r = 1 \frac{1}{2 - a^{r}} = \frac{n {.2}^{n - 1}}{2^{n} - 1} - 1$

Suggest Corrections

Similar questions

Q. If the equation

x^{n} - 1 = 0, n > 1, n \in N,

has roots

1, a_{1}, a_{2}, \dots, a_{n - 1},

then

Q. The equation

x^{n} = 1, n > 1, n \in N

has roots

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

. Then which of the following are correct?

Q. If

(1 + x)^{n} = n \sum r = 0 a_{r} x^{r} .

Then

(1 + \frac{a_{1}}{a_{0}}) (1 + \frac{a_{2}}{a_{1}}) \dots (1 + \frac{a_{n}}{a_{n - 1}})

is equal to

Q. Show that

\frac{1}{a_{0}} + \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} + \dots + \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}} = \frac{1}{a_{0} -} \frac{a_{0} x}{a_{1} + x -} \frac{a_{1} x}{a_{2} + x -} \dots \frac{a_{n - 1} x}{a_{n} + x}

Q. If

\frac{1}{\sqrt{2 x + 1}} {{(1 + \sqrt{2 x + 1})}^{n} - {(1 - \sqrt{2 x + 1})}^{n}} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{n} x^{n}

then

n

must be equal to

Join BYJU'S Learning Program

If xn−1=(x−1)(x−a)(x−a2)........(x−an−1), then prove that n−1∑r=112−ar=n2n−12n−1−1.

If $x^{n} - 1 = (x - 1) (x - a) (x - a^{2}) . . . . . . . . (x - a^{n - 1})$ , then prove that $n - 1 \sum r = 1 \frac{1}{2 - a_{r}} = \frac{n 2^{n - 1}}{2^{n} - 1} - 1$ .