Binomial expansion for approximation I've just had to do a homework on binomial expansion for approximation:

$1.07^9$

so:

$(1+0.07)^9$

To do binomial expansion you need a calculator for the combinations button (nCr), so why would use a more complicated method, which only gives an approximation be used over just typing 1.07^9 into a calculator?

(or is this never done in real life, and it's just a homework?)
Expanding the whole thing using Binomial Theorem gives you an exact value. Not an approximation.

To get an approximation you can consider a few terms from the expansion.

For instance, for "small" $x$, $1+nx$ is a "reasonable" approximation for $(1+x)^n$.

Notice that this corresponds to picking the first two terms from the binomial theorem expansion $(1+x)^n = 1 + \binom{n}{1} \ x + \binom{n}{2}\  x^2 + \dots + x^n$.

For example

$1.0007^9 \approx 1 + 9\times 0.0007 = 1.0063$ which agrees with $1.0007^9 = 1.0063176688422737867054812736724$ upto $4$ decimal places.

Depending on how accurate you want it, you could consider more terms from the binomial expansion.

This is based on the fact that for small $x$, as the power $r$ of $x$ gets larger, the term $x^r$ becomes small quite fast.
This looks like a badly thought-out problem to me. The first-order approximation is 1.63, but the true value is greater than 1.83. Can you give us the exact wording of the question?  

As for the practicality of this, I would say that being able to estimate such expressions is a useful skill in many branches of mathematics and statistics. Not essential, but useful.