nth row of Pascal's Triangle sum of the numbers is 2^n

We can try the Binomial Theorem here, since it has many binomial coefficients that are part of Pascal's Triangle,
$$\text{Binomial Theroem:}~(x+y)^n=\binom{n}{0}x^n+\binom{n}{1}x^{n1}y+\binom{n}{2}x^{n2}y^2+\dots +\binom{n}{n}y^n. $$We notice that the binomial coefficients
$$\binom{n}{0},\binom{n}{1},\dots,\binom{n}{n}, $$matchup with the nth row of Pascal's Triangle!
If we want to get rid of the variables, and want the something multiplied by the binomial coefficients to be 1, we can just let the variables to both be 1. So, we get
$$(1+1)^n=\boxed{2^n}. $$ 
Hi @professionalbronco!
Way to go, that's exactly how you would prove it algebraically!
Using the binomial theorem like this and plugging in \(1\) to get the coefficients is actually very useful in general.How about you give these problems a shot:
Find the sum of the coefficients of the expansion of \((x2y)^5\)
Find the sum of the coefficients of the expansion of \((x2y)^{12}\)
Find the sum of the coefficients of the expansion of \((x+yz)^2\)For some of them, you can just bash out, but try and find the shortcut!

@quacker88 Here are my solutions!
$$\binom{5}{0}+\binom{5}{1}+\binom{5}{2}+\dots+\binom{5}{4}+\binom{5}{5}=2^5=\boxed{32}. $$
P1 Solution: We notice that the sum of the coefficients, by the Binomial Theorem isP2 Solution: Similar to P1 Sol, the sum of the coefficients by the Binomial Theorem is
$$\binom{12}{0}+\binom{12}{1}+\binom{12}{2}+\dots+\binom{12}{11}+\binom{12}{12}=2^{12}=\boxed{4096}. $$However, I'm not sure about P3, because there are now 3 terms there. I could bash it out, but that doesn't seem very fun Should I write that as
$$[x+(yz)]^2? $$Then I can write it as
$$[x+(yz)]^2=1x^2+2x(yz)+1(yz)^2 $$so is the answer
$$\text{Sum of coefficients}=1+2+1=\boxed{4}? $$ 
@quacker88 I just noticed that S3 was incorrect. Here is a bash solution for the last one. We know that
$$(a+b+c)^2=a^2+2ab+2ac+b^2+2bc+c^2. $$So we can bash out this,
$$(x+yz)^2=x^2+2xy2xz+y^22yz+z^2 \Rightarrow \text{Sum of coefficients}=1+22+12+1=\boxed{1}. $$ 
Hey @professionalbronco !
Yeah that's a great catch for P3! And I really like the way you broke it up too. But there is actually a way to do it without expanding it out all the way. Let's take \((x2y)^3\) for example.Here's the thing: without expanding it, we can already tell that it will look something like \((ax^3+bx^2y+cxy^2+dy^3\)), does that make sense? We just have the terms, and the coefficients are \(a,b,c,d\). We might not know exactly what the coeffecients are, but we know that we're just looking for \(a+b+c+d\).
Wait, so if
\((x2y)^3=ax^3+bx^2y+cxy^2+dy^3\) and we're looking for \(a+b+c+d\), what do we need to plug in to just get the sum of the coefficients? Try using this tip and see if you can do the three problems faster.Actually, here's another thing to consider with P1 and P2. Expand \((x+2)^2\) and you should get \(x^2+4x+4\), right? But wait, binomial theorem says that the coefficients should be \(\binom20, \binom21, \binom22 \)...
$$\binom20\cdot(x)^2+\binom21\cdot(x)\cdot(2)+\binom22\cdot(2)^2 $$
Binomial theorem is actually still correct here. Notice how the equation is actuallySo binomial theorem definitely still works, but you just have to keep in mind that now we're multiplying \(x\)'s and \(2\)s now, instead of \(x\)'s and 1s.
Try and do the problems again with these tips now

@quacker88 Oh wait, we can just plug in 1 for the variables! S1: Plugging in x=1, y=1, we get
$$(12(1))^3=\boxed{1}. $$S2: Same thing, plugging in x=1,y=1, we get
$$(12(1))^{12}=\boxed{1}. $$S3: Plugging in (x,y,z)=(1,1,1), we get
$$(1+11)^2=\boxed{1}. $$
$$(ab+cd+ef+gh+\cdots)^3. $$Here are some of my favorites.(these are not valid problems BTW )
P1: Find the sum of the coefficients in the binomial expansionP2: Find the sum of the coefficients in the binomial expansion
$$(a2b+3c4d+5e6f+\cdots)^2. $$and
$$(a+2b+3c+4d+5e+6f+\cdots)^2. $$
P3: Find the sum of the coefficients in the binomial expansionNote: These go on forever.
Here are my solutions for these problems. (infinity is cool, isn't it?)S1: We can plug in all of them to be 1, and we get
$$(11+11+11+1\cdots)^3=\left(\frac{1}{2}\right)^3=\boxed{\frac{1}{8}}. $$See here to learn why 11+11+11+...=1/2.
S2: We can plug in all of them to be 1, and we get
$$(12+34+56+78+\cdots)^2=\left(\frac{1}{4}\right)^2=\boxed{\frac{1}{16}}. $$See here to learn why 12+34+56+...=1/4. (Stop at 2:34)
S3: Plugging in 1 for all of them, we get
$$(1+2+3+4+5+6+\cdots)^2=\left(\frac{1}{12}\right)^2=\boxed{\frac{1}{144}}. $$See here to learn why 1+2+3+4+5+6+...=1/12. (Start at 3:19, finish the video.)
Hope these problems were fun!

@professionalbronco Way to go, your answers for the three I gave you are really good!
And about the ones you talked about... I actually never thought about using it that way, that's a very interesting idea! I honestly can't process the idea of taking an infinite polynomial (with an infinite amount of variables), then actually squaring it. But in theory, I that should work exactly like you wrote it. This is why math is great! You can basically apply rules any way you want and they'll probably work. Way to go for coming up with this

@professionalbronco @quacker88 Sorry for the bump, but I finally figured out that the problems I shared were all wrong... the sums 11+11+11+... and 12+34+56+... and 1+2+3+4+5+6+... all diverge . All of them have applications of infinity which can result in weird results.

@professionalbronco No worries about the bump! Super glad to see you learning so much, and even coming back to correct an old post
You're absolutely right, these sums don't converge, so you can't manipulate them as if they were actual equations. In fact, \(1+2+3+4+...=\frac{1}{12}\) is just really misleading in general. While \(\frac{1}{12}\) is related to it, it's not really "equal" to it in the normal sense.
Great job once again!