How can we solve for x for the equation below without trial and error?

Question

A = 1/(1+x) + 1/(1+x)^2+1/(1+x)^3+...+1/(1+x)^n

where A is a constant, n is the highest degree and ^n denotes an exponent raised to the nth power

1/(1+x... actually stands for 1/(1+x)^n

Answer 1

a=1/(1+x)
r=1/(1+x)
Sn=(1/1+x)*{1/1+x)^n -1}/{1/(1+x)-1}

Answer 2

What you have here is a geometric series-
Let 'r' = 1/(1 + x).
Then A = r + r^2 + r^3 + ...

I'm going to assume you are familiar with the formula for such an infinite geometric series (it should be in your textbook). The series would start at n=0, so we'll add r^(0) to both sides:
A + 1 = r^0 + r^1 + r^2 + r^3 + ...

By the formula, we have:
A + 1 = 1/(1-r).

Replace 'r' with 1/(1+x):
A + 1 = 1/(1-1/(1+x)).

Now a little algebra:
A+1 = 1/( (1+x)/(1+x) - (1)/(1+x) ),
A+1 = 1/((1+x-1)/(1+x)),
A+1 = 1/(x/(1+x)),
A+1 = (1+x)/x,
A+1 = 1/x + x/x,
A+1 = 1/x + 1,
A = 1/x,
x = 1/A.

Answer 3

try factoring out all the x..