Binomial expansion, how to do them quickly? I'm currently preparing for a test where I'm bound to do a couple of binomial expansions. Since I never encountered them in my formal education, I looked how to do them myself and found out:
$ (x+y)^n = \displaystyle \sum_{k=0}^n \binom{n}{k} x^k y^{n-k} $ 
I'm using this formula for my binomial expansion, but this takes a long time, especially to calculate the binomial coefficient (we aren't allowed to use a calculator).
Consider the following (easy) example: Write down the binomial expansion of $(4+x)^4$, and hence evaluate $(4.2)^2$ to 2 decimal places.
I use the above formula  to get $256 + 256x + 96x^2+16x^3 + x^4$. Fill in for $x=0.2$.


*

*Here I encounter my first problems: How do you know how many terms to use when you are told to evaluate to $n$ decimal places?

*Unrelated to the title: $(4.2)^4 = 311.1696$. When you use 4 terms you get $311.168$. Would the correct answer be $311.16...$ or $\approx 311.17$?

*Are there other, quicker ways of doing binomial expansions, which can rid me of the problem of binomial coefficients?
 A: Suppose I wanted to expand $(x + 4)^4$. Using the above formula, $x = x$ and $y = 4$. So 
$${4\choose0}x^44^0 + {4\choose1}x^34^1 + {4\choose2}x^24^2 + {4\choose3}x^14^3 + {4\choose4}x^04^4$$
Knowing Pascal's triangle, I can use $1, 4, 6, 4, 1$ as my coefficients. So the answer is 
$$x^4 + 16x^3 + 96x^2 + 256x + 256.$$

Here I encounter my first problems: How do you know how many terms to use when you are told to evaluate to n decimal places?

In terms of $x$, you go down from $x^n$ to $x^0.$ I feel as if I misunderstood this question, however.

Unrelated to the title: (4.2)4=311.1696. When you use 4 terms you get 311.168. Would the correct answer be 311.16... or ≈311.17?

It doesn't really matter here, but when rounding, it's 311.17. 168 is over 165, which rounds t0 170.

Are there other, quicker ways of doing binomial expansions, which can rid me of the problem of binomial coefficients?

Not that I know of.
A: Sorry if there is a rule against posting on questions that have not been active lately, feel free to delete if there is, but I thought this method might be helpful for someone else who doesn't want to look at Pascal's triangle or won't be given it in a test.
Here is a well-known alternative method you might find faster. Let
$$\binom{n}{k}:=m_k.$$ Then
$$\frac{m_{k+1}}{m_k}=\frac{k!\left(n-k\right)!}{\left(k+1\right)!\left(n-k-1\right)!}=\frac{n-k}{k+1}.$$ This means if you know one coefficient you can calculate all the rest, by multiplying the first coefficient $m_k$ by the corresponding index $n-k,$ then dividing by the number of terms you have in the sum so far, $k+1.$ So for example, if you have $$(a+b)^n=\sum\limits_{k=0}^{n}\binom{n}{k}a^{n-k}b^k,$$ then if we take $n=37,$ you know $(a+b)^{37}=a^{37}+\cdots,$ so your first coefficient is $1.$ It corresponds to the index $n-k=37,$ and you have one term so far. So, your next coefficient is
$$1\times\frac{37}{1}=37.$$ Next, $37\times36/2=666.$ The fourth coefficient is $666\times35/3=7770,$ getting $$(a+b)^{37}=a^{37}+37a^{36}b+666a^{35}b^2+7770a^{34}b^3+\cdots$$and so on until you get half of them and then use the symmetrical nature of the binomial theorem to write down the other half.
