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

sciencesolve | Certified Educator

You need to test if the following quotients are equal, such that:

`(x^2 - 1)/(x + 1) = (x^3 - x^2 - x + 1)/(x^2 - 1) = q`

You need to convet the difference of squares into a product, such that:

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

You need to group the terms of `x^3 - x^2 - x + 1` such that:

`(x^3 - x^2) - (x - 1)`

Factoring out x^2 in first group yields:

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

Factoring out x - 1 yields:

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

You need to write the factored form of quotients such that:

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

Reducing duplicate factors yields:

`x - 1 = x - 1`

Hence, testing if the relation between quotients `(x^2 - 1)/(x + 1)` and `(x^3 - x^2 - x + 1)/(x^2- 1)` is valid yields `(x^2 - 1)/(x + 1) = (x^3 - x^2 - x + 1)/(x^2- 1)` , thus, `x+1, x^2-1, x^3-x^2-x+1` are the successive members of geometric series.

giorgiana1976 | Student

For a geometric sequence, the middle term, that is  x^2-1, must 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'll factorize last term x^3-x^2-x+1, we'll get:

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

We'll factorize one more time:

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)

Since LHS=RHS, therefore the given terms are the consecutive terms of a geometric progression.