Sum of all of the numbers in row n of Pascal’s triangle? Explain why this happens, in terms of the way the triangle is formed... Just to clarify there are two questions that need to be answered:
1)Explain why this happens, in terms of the way the triangle is formed.
2) Explain why this happens,in terms of the fact that the combination numbers count subsets of a set.
I know the sum of the rows is equal to $2^{n}$. However I am stuck on the other questions. I also have to assume I don't know the binomial theorem just yet.
I thought my explanation would be just add up the numbers in the rows, but I'm not sure if that is even what I need to do.
Please help. Thank you!
 A: For any subset of a set of $n$ elements, and for any element of that set, the subset either contains that element or not. Hence, there are $2^n$ different subsets for any set of $n$ elements.
This is of course the sum, with $i$ ranging from $0$ to $n$, of the number of different subsets with $i$ elements, of which there are $n \choose i$ ... Which we know as the $i$-th entry of the $n$-th row in the triangle.
So: $\sum_{i=0}^n {n \choose i} =2^n$, and therefore the sum of all entries in row $n$ equals $2^n$
A: Every element in row $n$ contributes to exactly two elements in row $n+1.$
A: *

*Notice the first row has sum $1=2^{0}$, every other row is recursively defined to be the previous row added to itself under the shift map which is a bijection hence the sum of every row $r_{n+1}=2r_{n}$ where $r_{i}$ is the $i^{th}$ row. Solving the initial condition and the recurrence relation yield that the sum of every row is $2^{n}$.

*In terms of subsets the sum of the $n^{th}$ row is the sum of all subsets of size $k<n$. This is the same as all subsets of n since a subset of n can not have $n+1$ elements hence $2^{n}$ from elementary set theory.
A: Use this problem as a gateway to learn one of the central tricks in all of Mathematics: Counting things in different ways to obtain a result that may not be obvious otherwise.
Say you have $n$ light switches in a row. Let's count the total number of different on/off configurations. On the one hand, each switch can be on or off, so there are $2^n$ configurations.
But we can count differently, grouping configurations in categories; namely, we can count how many configurations have $0$ on switches, then how mny have $1$ on switch, then how many have $2$, and so on, until we count how many configurations have $n$ on switches.
The consecutive answers are $\binom{n}{0}$, $\binom{n}{1}$, $\binom{n}{2},\ldots,\binom{n}{n}$, so the equality is proved. More to the point, the equality is explained as coming from counting something in two different ways.
The punchline is that you can understand the equality as encoding something about the internal structure of configurations of $n$ switches. Hopefully you find such notion inspiring.
A: In case you already know that the entries in Pascal's triangle are the binomial coefficients, i.e., that the $k$th entry in the $n$th row, $n\choose k$, is the coefficient of $x^k$ in the expansion of the binomial $(1+x)^n$, then the sum of these coefficients is simply the evaluation at $x=1$, i.e., 
$$ \sum_{k=0}^n{n\choose k}=\sum_{k=0}^n{n\choose k}1^k=(1+1)^n=2^n.$$
A: If $a^n_k$ is the $k$th element of the $n$th row, then the triangle is defined by $a^0_0=1$ and $a^{n+1}_k=a^n_k+a^n_{k-1}$ for $n\ge 1$ (we also need to specify that $a^n_k=0$ for $k<0$ or $k>n$).
Then we can compute the sum of the elements in row $n+1$ by $$\sum_k a^{n+1}_k=\sum_k(a^n_k+a^n_{k-1})=\sum_ka^n_k+\sum_ka^n_{k-1}=2\sum_ka^n_k$$
Combine this with the fact that the sum of the terms in the first row is $1$, and this shows that the sum of the $n$th row is $2^n$.
Of course, I've been sloppy here by not including the bounds of the sum and not keeping track of what happens around $k=0$ and $k=n+1$, but a little more attention to detail will make this into a rigorous proof of the identity.
