Is there a trick for adding and subtracting binomial coefficients?

Question

I had a problem for which I needed to evaluate: $$\sum_{n=1}^3{5\choose n}$$ I understand the basic concept of binomial coefficients, but I haven't been able to find an easy way to evaluate this by hand simply based on the numbers involved. Is there a trick I can use for adding and subtracting binomial coefficients so I don't have to use Wolfram Alpha every time I want to do something similar?

Edit: I cannot evaluate coefficients myself. That's why I'm looking for a trick to add using only the numbers displayed.

Answer 1

$5 \choose 1$ is just $5$. There are two more terms in the sum. Can you evaluate ${5 \choose 2}$ and $5 \choose 3$?. Hint: they are equal. Then just add them up.

Answer 2

Since

$\begin{array}\\ \dfrac{\binom{m}{n+1}}{\binom{m}{n}} &=\frac{\frac{m!}{(n+1)!(m-n-1)!}}{\frac{m!}{n!(m-n)!}}\\ &=\frac{m!n!(m-n)!}{m!(n+1)!(m-n-1)!}\\ &=\frac{m-n}{n+1}\\ \end{array} $

we have $\binom{m}{n+1} =\frac{m-n}{n+1}\binom{m}{n} $.

Starting with $\binom{m}{0} =1$ we get

$n=0, \binom{m}{1} =\frac{m}{1}\binom{m}{0} =m $,

$n=1, \binom{m}{2} =\frac{m-1}{2}\binom{m}{1} =\frac{m-1}{2}m =\frac{m(m-1)}{2} $,

$n=2, \binom{m}{3} =\frac{m-2}{3}\binom{m}{2} =\frac{m-2}{3}\frac{m(m-1)}{2} =\frac{m(m-1)(m-2)}{6} $,

and so on.

At each stage, the fraction you multiply by has its numerator decreased and denominator increased.