If x is positive then the sum to infinity of the series
\(\frac{1}{1+3 x}\) - \(\frac{1-3 x}{(1+3 x)^2}\) + \(\frac{(1-3 x)^2}{(1+3 x)^3}\) - \(\frac{(1-3 x)^3}{(1+3 x)^4}\) ..... ∞ is
1
\(\frac{1}{2}\)
2
\(\frac{1}{6x}\)
3
\(\frac{1}{6x(1+3x)}\)
4
\(\frac{1}{2(1+3x)}\)