The solution is `x=-1/2 - sqrt(5)/2`

We are asked to solve `1+((x+1)/(x-1))+((x+1)/(x-1))^2+...=x^2/2`

Normally we cannot add an infinite number of terms, but we certainly can if they form a geometric progression (or geometric series.) In order for an infinite geometric series to have a sum, the common ratio r must be less than 1 in absolute value.

Each term is found by multiplying the previous term by `(x+1)/(x-1)` which is the common ratio r. As long as `|(x+1)/(x-1)|<1` the series has a finite sum.

The sum of an infinite geometric series is given by `S=a/(1-r), |r|<1` where S is the sum, a is the first term, and r is the common ratio as discussed. Here we want the sum to be `x^2/2` so we get:

`1/(1-(x+1)/(x-1))=x^2/2` Multiply the left hand side by `(x-1)/(x-1)` to get

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

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

Then, `x^2=1-x ==> x^2+x-1=0`.

We can use the quadratic formula (or completing the square) to find the value(s) of x that are solutions:

`x=(-1 +- sqrt(1^2-4(1)(-1)))/(2(1))`

`x=(-1+sqrt(5))/2" or " (-1-sqrt(5))/2`

With the restriction that `|(x+1)/(x-1)|<1` the solution is `x=-1/2-sqrt(5)/2`