Binomial expansion, how to do them quickly?

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.

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.