If you have a look to my answer to this question, you will see a "rather simple" approximation.
Using your numbers, my notations and whole numbers $$a=\frac {100}{1335}=\frac {20}{267}\qquad \text{and}\qquad n=20 \implies y=\frac{38}{801}$$ Computing the coefficients
$$c_1=-\frac{59917}{6}\qquad c_2=-\frac{2203844}{21}\qquad c_3=-\frac{189356849}{945}$$
$$d_1=-9983\qquad d_2=-\frac{34418015}{252}\qquad d_3=-\frac{393776177}{945}$$
Using the formula, you get
$$i=\frac{6739162539251212}{160365841116253677}\approx 0.04202368$$ while the "exact" solution would be
$$i\approx 0.04202375$$
This could easily be implemented in any programmable calculator.
Edit
From a programming point of view, it could be more convenient to rewrite my previous formulae as
$$r=y\,\frac{\sum_{k=0}^3 c_k\, y^k }{\sum_{k=0}^3 d_k\, y^k }\qquad \text{where}\qquad y=\frac{2 (a n-1)}{n+1} $$ with
$$\left(
\begin{array}{cc}
\color{red} {k} & \color{red} {c_k} \\
0 & 15120 \left(n^2-20 n+1\right) \\
1 & -1260 (n+2) \left(8 n^2+113 n-13\right) \\
2 & -90 \left(90 n^4+355 n^3+942 n^2+705 n-148\right) \\
3 & -4 (n+2) \left(194 n^4+388 n^3+687 n^2+493 n-142\right)
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
\color{red} {k} & \color{red} {d_k} \\
0 & 15120 \left(n^2-20 n+1\right) \\
1 & -3780 \left(2 n^3+57 n^2+57 n-8\right) \\
2 & -60 \left(170 n^4+823 n^3+1560 n^2+508 n-145\right) \\
3 & -(2 n+1) \left(874 n^4+1559 n^3+3336 n^2+1139 n-428\right)
\end{array}
\right)$$
