Super fraction #1

Let $g(n) = f(n) - 1$ . Then $g(n) = \frac{1}{n+g\left(n\right)}$ . Solving for $g(n)$ :

$g\left(n\right)^{2}+ng\left(n\right)-1=0$

$g\left(n\right)=\frac{-n \pm \sqrt{n^{2}+4}}{2}$

Substituting $g(n) = f(n) - 1$ and solving for $f(n)$ yields:

$f\left(n\right)=\frac{-n+2 \pm \sqrt{n^{2}+4}}{2}$

Therefore, $\boxed{a = -1, b = -2, c = 4, d = \frac{1}{2}}$

$f(n)=1+\dfrac{1}{n-1+f(n)}$ $f(n)(n-1+f(n))=n-1+f(n)+1$ $f(n)^{2}+(n-1)f(n)=n+f(n)$ $f(n)^{2}+(n-2)f(n)-n=0$ $f(n)=\dfrac{-n+2\pm(n^{2}-4n+4+4n)^{\frac{1}{2}}}{2}$ $\begin{cases} a=-1 \\ b=-2 \\ c=4 \\ d=0.5 \end{cases}$ $a+b+c+d=\boxed{1.5}$