Using xn+1=3.6xn2+2.1\displaystyle x_{n+1} = \frac{3.6}{x_n^2} + 2.1xn+1=xn23.6+2.1 with x0=2x_0 = 2x0=2
Find the values of x1x_1x1, x2 x_2\,x2 and x3x_3x3.
Explain the relationship between the values of x1x_1x1, x2 x_2\,x2 and x3 x_3\,x3 and the equation x3−2.1x2−3.6=0x^3 - 2.1x^2 - 3.6 = 0x3−2.1x2−3.6=0
124 exam-style questions on OCR GCSE Maths Iteration. Each one has a worked solution and a mark scheme showing where the marks go.