By choosing a suitable binomial expansion of the form $(1±kx)^n$ and a suitable value of x. Find $(1.01)^{20}$ correct to 3 s.f.

Question

I have tried replacing kx by 0.1, square rooting 0.1 and squaring 0.1, but none of these methods worked. I don’t know where to start from. I would appreciate some assistance on this question. Thanks

Answer 1

If you write out the binomial expansion $$1.01^{20}=1^{20}+{20 \choose 1}1^{19}\cdot 0.01+{20 \choose 2}1^{18}\cdot 0.01^2 \ldots$$ You can note that the binomial increases by at most a factor $20$ while the power of $0.01$ divides by $100$. You can bound the tail by a geometric series of ratio $\frac 15$, so the tail is less than $\frac 14$ of the last term you include. The first two terms give $1.2$, so once a term is less than $0.01$ you can stop.

Answer 2

The accepted answer has a major flaw in thinking that, if a real number $s$ and "true" value $S$ differs by less than 0.01 then they both round to the same 2dp numbers. But this is easily shown to be false, e.g., $1.005-10^{-12}$ and $1.005$ differs by $10^{-12}\ll 0.01$ but rounds to different numbers to 2dp (or 3sf). In other words,

Calculate $x$ to within error smaller than $10^{-n}$ (or half that).

and

Calculate $x$ to n decimal places

are different problems.

So you need to keep track of both the current partial sum and potential error. Only if you can guarantee everything in range round to the same decimal number can you stop.

So:

1. First term 1, error $0<\varepsilon<\frac14$, potential range $(1,1.25)$ is too broad.

2. Two term sum $1+\binom{20}{1}\cdot 0.01=1.20$, error $0<\varepsilon<\frac14\cdot 0.2=0.05$, potential range $(1.20,1.25)$ again too broad

3. Three term sum $1+\binom{20}{1}\cdot 0.01+\binom{20}{2}\cdot 0.01^2=1.219$, error $0<\varepsilon<\frac14\cdot 0.019=0.00475$, potential range $(1.219, 1.22375)$ falls under what is represented by 1.22 to 3sf.