Dear Uncle Colin,

I’m practicing for the Oxford PAT and have been asked how many terms of the binomial expansion would be needed to determine $(3.12)^5$ to one decimal place? I don’t really know where to start.

- Knows Expansions (Binomial); Lacks Explanations

Hi, KEBLE, and thanks for your message!

That’s an odd question, but it’s easy enough to work out $\br{3+0.12}^5$ using the binomial expansion:

$1 \times 3^5 \times 0.12^0 = 243$; $5 \times 3^4 \times 0.12^1 = 81 \times 0.6$; $10 \times 3^3 \times 0.12^2 = 27 \times 0.144$; $10 \times 3^2 \times 0.12^3 = 9 \times 0.01728$; $5 \times3^1 \times 0.12^4 \approx 3 \times 0.002$; $1 \times 3^0 \times 0.12^5 \approx 0.0002$.

The fourth term (with $0.12^3$) is about 0.1, which would certainly change the first decimal place; the fifth term is about 0.006, which is likely to change the second (but unlikely to change the first).

That means, four terms are required to get an answer you’re reasonably confident is correct to 1dp.

Hope that helps!

- Uncle Colin