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Use the formula $x_{n+1} = \frac{(x_n)^3}{30} + 2$ with $x_1 = 2$ to calculate $x_2$ and $x_3$ - OCR - GCSE Maths - Question 15 - 2018 - Paper 6

Question 15

$Use-the-formula-$x_{n+1}-=-\frac{(x_n)^3}{30}-+-2$-with-$x_1-=-2$-to-calculate-$x_2$-and-$x_3$-OCR-GCSE Maths-Question 15-2018-Paper 6.png$

Use the formula $x_{n+1} = \frac{(x_n)^3}{30} + 2$ with $x_1 = 2$ to calculate $x_2$ and $x_3$. Round your answers correct to 4 decimal places.

Worked Solution & Example Answer:Use the formula $x_{n+1} = \frac{(x_n)^3}{30} + 2$ with $x_1 = 2$ to calculate $x_2$ and $x_3$ - OCR - GCSE Maths - Question 15 - 2018 - Paper 6

Step 1

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Answer

Substituting $x_1 = 2$ into the formula:

$x_2 = \frac{(2)^3}{30} + 2$

Calculating gives:

$x_2 = \frac{8}{30} + 2 = 0.2667 + 2 = 2.2667$

Therefore, $x_2 = 2.2667$ .

Step 2

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Answer

Using the value of $x_2$ calculated above:

$x_3 = \frac{(x_2)^3}{30} + 2 = \frac{(2.2667)^3}{30} + 2$

Calculating $(2.2667)^3$ gives approximately $11.6340$ , so:

$x_3 = \frac{11.6340}{30} + 2 = 0.3878 + 2 = 2.3878$

Therefore, $x_3 = 2.3878$ .

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