Question 1185530
{{{"f'(x)" = 2*(1+2x)^(-1)}}}
{{{"f''(x)" = (-1)^1*2^2*(1+2x)^(-2)}}}
{{{"f'''(x)" = (-1)^2*2^3*2!*(1+2x)^(-3)}}}
....
Inductively, {{{f^((k))(x) = (-1)^(k-1)*2^k*(k-1)!*(1+2x)^(-k)}}}, and so 

{{{f^((k))(0) = (-1)^(k-1)*2^k*(k-1)!}}}


===> {{{ln(1+2x) = sum( (f^((k))(0)/k!)*x^k, k=0,infinity) = sum( (((-1)^(k-1)*2^k*(k-1)!)/k!)*x^k, k=1, infinity) = red(sum( (((-1)^(k-1)*2^k)/k)*x^k, k=1, infinity)) }}}