Inspired Infinite Equation

The RHS of the equation is a geometric series with first term $x^2$ and common ratio $-x$ . Its sum is $\frac{x^2}{1+x}$ , for $|x|<1$ . Our equation thus becomes $\frac{x^2}{1+x}=1$ , or $x^2=1+x$ . Solving using the quadratic formula, the roots are $\frac{1-\sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}$ , the latter of which can be discarded as it is greater than $1$ . Thus, $\frac{1-\sqrt{5}}{2}$ is our only root, so $a+b+c+d=1-1+5+2=\boxed{7}$ .

$\begin{aligned} 1 & = x^2 - x^3 + x^4 - x^5 + ... \\ 1 & = x^2 -x(x^2 - x^3 + x^4 - x^5 + ...) \\ 1 & = x^2 - x(1) \end{aligned}$

$\begin{aligned}\Rightarrow x^2 - x - 1 & = 0 \\ x & = \dfrac{1\pm \sqrt{5}}{2} \\ \Rightarrow x & = \dfrac{1- \sqrt{5}}{2} \quad \quad \small \color{#3D99F6} {[\text{Since }x < 1]} \end{aligned}$

$\Rightarrow a + b + c + d = 1 - 1 + 5 + 2 = \boxed{7}$

$\displaystyle \sum _{k=0}^\infty a*r^k=\dfrac a {1-r} ~iff~|r|<1\\a=x^2,~~r=-x.\\\therefore1=\dfrac{ x^2}{1+x}\\ x^2-x-1=0.\\ x=\dfrac{1\pm \sqrt{1-4*(-1) } } 2\\ For~x<|1|~~x= \dfrac{1-\sqrt{1-4*(-1) } }2\\ \therefore~\dfrac{a + b\sqrt c} d=\dfrac{1-1*\sqrt 5} 2\\a+b+c+d=1 - 1+ 5+ 2= ~~~~\Large \color{#D61F06}{ 7}$

1=x²-x³+x⁴+...
1=x²(1-x+(x²-x³+...))
1=x²(1-x+1)
2x²-x³-1=0 (x-1)(x²-x-1)=0 the roots of the equation are X=1 or x=½(1-√5) or x=½(1+√5) The serie converge only if 0<x<1 so x must be x=½(1-√5) and the answer was 7

We get the golden ratio when solving for $x$ but don't forget to reject $\frac{1+\sqrt{5}}{2}$ as $-1<x<1$ for the sum to infinity formula to hold.