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A sequence of numbers $x_1, x_2, x_3, \ldots$ , starts with $x_1 = 2$, and, if you know any term $x_n$, you can find the next term $x_{n+1}$ using the formula: $$x_{n+1} = \frac{1}{2}\bigl(x_n + \frac{3}{x_n}\bigr)$$ Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence?
Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give?
What happens when you try this method for fourth roots or fifth roots etc.?
Route to Root
Age 16 to 18
Challenge Level 
A sequence of numbers $x_1, x_2, x_3, \ldots$ , starts with $x_1 = 2$, and, if you know any term $x_n$, you can find the next term $x_{n+1}$ using the formula: $$x_{n+1} = \frac{1}{2}\bigl(x_n + \frac{3}{x_n}\bigr)$$ Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence?
Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give?
What happens when you try this method for fourth roots or fifth roots etc.?