Sum of coefficients in binomial theory.

Question

While trying to get introduced to binomial theory at university's website, I learned about the sum of binomial coefficients, and they showed me some of the features, and one of them was the pyramid of coefficients sum which is:

$$1 = 1$$

$$1+1 = 2$$

$$1+2+1 = 4$$

$$1+3+3+1 = 8$$

$$1+4+6+4+1 = 16$$

and then they asked me to filled the blank with something that describes the sum:

$$\sum_{i=0}^n \binom{n}{i} = \left[ \phantom{\frac 1 1} \right]$$

If I understood right what they want, they how do I describe the sum ?

Answer 1

There are a variety of different ways to understand this. One of those is the combinatorial way of understanding it, which says that binomial coefficients count subsets of specified sizes (i.e. "combinations") and the sum of all the coefficients in one row counts all subsets of all sizes, of a set of a specified size.

That is worth understanding, but I'll give a quicker answer here. In constructing Pascal's triangle, each entry besides all the $1$s is the sum of the two entries above it. That means every number in each row gets added into the previous row twice. That means the sum of the entries in the next row will be twice as big. Every time you got down one row, the sum gets multiplied by exactly $2$.

Answer 2

So you need

$$\sum_{i=0}^n {n \choose i}.$$

For $n=4$ this is

$$\sum_{i=0}^4 {4 \choose i} = {4 \choose 0} + {4 \choose 1} + {4 \choose 2} + {4 \choose 3} + {4 \choose 4} = 1 + 4 + 6 + 4 + 1 = 16.$$

If you can see the power-ful pattern in the series in your question, then hopefully this will get you two (uhhh, to) the correct answer. (Note the hints in italics.)

Answer 3

Note that Pascal's triangle is completed by adding pairs of elements of row $n-1$ (including often unnoticed zeros) to obtain row $n$ and this can be used to prove that row $n$ has twice the sum of row $n-1$.

Alternatively consider the binomial expansion of $(1+1)^n$ for a direct proof.