How to Find the Sum of Roots of the Polynomial \( x^{100} - 3x + 2 = 0 \)?

[Cade]

How would I go about approaching this problem?

Given the polynomial:
x^100 - 3x + 2 = 0

Find the sum 1 + x + x^2 + ... + x^99 for each possible value of x.

 

[DonAntonio]

How would I go about approaching this problem?

Given the polynomial:
x^100 - 3x + 2 = 0

Find the sum 1 + x + x^2 + ... + x^99 for each possible value of x.

If you meant that x is a root of the polynomial X^{100}-3X+2 , then
1+x+...+x^{99}=\frac{x^{100}-1}{x-1}=\frac{3x-3}{x-1}=3

DonAntonio

 

[Cade]

Interesting, thanks, how did you derive that?

 

[DonAntonio]

Interesting, thanks, how did you derive that?

First equality: sum of a geometric sequence.

Second equality: x^{100}-3x+2=0\Longrightarrow x^{100}=3x-2

Third equality: trivial algebra

DonAntonio

 

[Cade]

Oh, I didn't realize the first part was the sum of a geometric series. Thanks for your help.

 

[coolul007]

isn't a trivial solution to the equation equal to 1, then then sum would be greater than 3, This is the solution that makes the geometric sum equation impossible as you are dividing by zero.

 

[DonAntonio]

isn't a trivial solution to the equation equal to 1, then then sum would be greater than 3, This is the solution that makes the geometric sum equation impossible as you are dividing by zero.

Indeed. So for \,\,x=1\,\,,\,\,1+1^1+1^2+...+1^{99}=100\,\, , and for all the other roots it is what I wrote before.

Thanx.

DonAntonio