Using xn+1=15xn2+4\displaystyle x_{n+1} = \frac{15}{x_n^2 + 4}xn+1=xn2+415 with x0=1x_0 = 1x0=1
Find the values of x1x_1x1, x2 x_2\,x2 and x3x_3x3.
40 exam-style questions on Edexcel GCSE Maths Iteration. Each one has a worked solution and a mark scheme showing where the marks go.