(i) by substituting x=2/3, use the series expansion of (1-x)/sqroot(4+3x) in ascending powers of x, up to and including the term in x^3, to find an approximate value of sqroot6 in the form of a fraction.
(ii) state the range of values of x for which the series expansion is valid.
(iii) deduce the equation of the tangent to the curve y= (1-x)/sqroot(4+3x) at the point where x=0
(1-x)/sqrt(4+3x) = 1/2 - 11/16 x + 75/256 x^2 - 351/2048 x^3 + ...
so the tangent line is y = 1/2 - 11/16 x