I am trying to solve a 2 part question, the first part is to expand $\sqrt{1-x}$ up to and including the term $x^2$ which I did.
This gives me $1-(0.5)x-(0.125)x^2 + ...$
However, the 2nd part of the question says that by taking $x=(1/64)$ in the expansion of part (i) -> which refers to the one above, deduce that $\sqrt 7 = 10837/4096$. How do i do this part?
EDIT (After Hint Given By SimpliFire):
$$ \sqrt{1-\frac1{64}}=\frac{\sqrt{63}}8=\frac38\sqrt7 $$ $$ =\frac83\sqrt{1-\frac1{64}}\;=\sqrt7 $$ $$ =\frac83\left(1-\frac12\left(\frac1{64}\right)-\frac18\left(\frac1{64}\right)^2\right)\; $$ $$ =\frac83\left(1-\frac1{128}-\frac1{32768}\right)\; $$ $$ =\frac83\left(\frac{32768}{32768}-\frac{256}{32768}-\frac1{32768}\right)\; $$ $$ =\frac83\left(\frac{32511}{32768}\right)\; $$ $$ =\frac{260088}{98304} $$ $$ \frac{10837}{4096}\;\approx\sqrt7 $$
I'm not sure if I can present it like this, if it is okay i will post it as my answer :X
Correct approach!
Some minor notational errors. Here's how I would lay it out:
Since
$$ \sqrt{1-\frac1{64}}=\frac{\sqrt{63}}8=\frac38\sqrt7 \implies\frac83\sqrt{1-\frac1{64}}=\sqrt7,$$ from the previous part, we have the approximation $$\begin{align}\sqrt7\approx\frac83\left(1-\frac12\left(\frac1{64}\right)-\frac18\left(\frac1{64}\right)^2\right)&=\frac83\left(1-\frac1{128}-\frac1{32768}\right) \\&=\frac83\left(\frac{32768}{32768}-\frac{256}{32768}-\frac1{32768}\right)\\&=\frac83\left(\frac{32511}{32768}\right)=\frac{260088}{98304}=\frac{10837}{4096}\end{align}$$ as required.