# Prove that the terms x+1, x^2-1, x^3-x^2-x+1 are the terms of a G.P.

hala718 | High School Teacher | (Level 1) Educator Emeritus

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x+ 1. x^2 -1 . x^3 - x^2 -x +1

If the terms are parts of a G.P. then

(x^2 -1) / (x+1) = (x^3 -x^2 -x + 1)/(x^2 -1) = R

Ley us simplify:

We know that:

x^2 - 1 = (x-1)(x+1)

x^3 - x^2 - x + 1= (x^2(x-1) - (x-1) = (x-1)(x^2-1)= (x-1)^2*(x+1)

Now Let us substitute:

==> (x-1)(x+1)/(x+1) = (x-1)^2 (x+1)/(x-1)(x+1)

==> (x-1) = (x-1) = R

Then the terms. are parts. of a G.P.

neela | High School Teacher | (Level 3) Valedictorian

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Any consecutive terms of GP should bear the same ratio.

We notice that the second term divided by 1st term gives us  (x^2-1)/(x-1) = (x+1))(x-1)/(x-1) = x+1.

The third term divided by the 2nd term = (x^3-x^2-x+1)/(x^2-1).

x^2-1) x^3  -x^2 - x+1( x

x^3 -0x^2 - x

-------------------------------

x^2-1)x^2   + 1( x -1

x^2   +1

-------------------------------

0.

Therefore  (x^3-x^2-x+1)/(x^2-1) = x+1.

Thus the common ratio is x+1  between the consecutive terms and the starting term is x-1.

Therfore x+1 , x^2 -1 , x^3 -x^2-x+1  is in GP.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To prove that the given terms are the consecutive terms of a geometric series, we'll use the constraint imposed to the middle term of 3 consecutive terms of a geometric series.

The middle terms is  x^2-1 and it has to be the geometric mean of the neighbor terms.

x^2-1 = sqrt[(x+1)*(x^3-x^2-x+1)] (1)

We notice that if we factorize last term x^3-x^2-x+1, we'll get:

x^2(x-1) - (x-1)

We'll factorize again:

x^2(x-1) - (x-1) = (x-1)(x^2 - 1) (2)

The factor x^2 - 1 is a difference of squares:

x^2 - 1 = (x-1)(x+1) (3)

We'll substitute (3) in (2):

x^2(x-1) - (x-1) = (x-1)(x-1)(x+1)

x^2(x-1) - (x-1) = (x-1)^2*(x+1) (4)

We'll substitute (4) in (1):

x^2-1 = sqrt[(x+1)*(x-1)^2*(x+1)]

x^2-1 = sqrt[(x+1)^2*(x-1)^2]

x^2-1 = (x+1)(x-1)

We have obtained an identity, so the given terms are the consecutive terms of a geometric progression.