+202 Let's say we have \((x+1)^n\). What will be the variable of the term with coefficient \(\binom{n}{k}\)?
+343 n-k+1? I just guessed. I used the pascals triangle, not sure if this is right....
+658 Never learned pascals triangle. So I'm not sure what the question means.
By subsituting real numbers (specifically integers)
If n = 2
x^2 + 2x + 1 and n choose k will be 2
then the variable will be x
If n = 3
(x + 1)(x^2 +2x + 1)
x^3 + 2x^2 + x + x^2 + 2x + 1
x^3 + 3x^2 + 3x + 1
n choose k will be 3
then the variable will be x
So far I think the answer is x
+202 I'm fairly certain there's no way that can be the answer. Sorry. And anyway, Pascal's Triangle is really interesting and you should check it out. There are countless things you can do with it. In this case, Pascal's Triangle's numbers are also combinations, which is really cool. This means you can use it as a way to find the coefficients in the Binomial Theorem, although 9 times out of 10 you're better off just computing the combination.
+658 I just searched it up, and it is so useful! Maybe I should have another attempt at this problem.
+658 I am kind of confused the purpose for the \(k\) variable.
\(\frac{n!}{k!(n-k)!}\)
Whatever \(n\) is, the expansion has the coefficients listed in the \((n+1)\)th row of Pascal's Triangle.
+658 Woah! Never touched summation before! I did some reasearch from this website: https://medium.com/i-math/the-binomial-theorem-explained-6464f41e5268
Let us subsitute in known values
1. \((x+1)^n=\sum\limits_{k=0}^n {n \choose k}x^{n-k}1^k\)
2. \((x+1)^n=\sum\limits_{k=0}^n {n \choose k}x^{n-k}\)
I realized that any expansion of (x + 1)n will have coefficients in the nth row in pascal's triangle. I think that means the variable is x(n-k).
....
So in other words I have no idea but for some reason x(n-k) is working.
+202 I believe you are right. And summation is an amazing tool as well. You should expose yourself to it.
+658 Yay ! 
I will be sure to learn summation. It will be nice to be ahead of my class next year 
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+202 If you really want to be ahead of the curve, take advantage of AoPS classes. I'm recommending them for their level of depth.
+658 My parents won't allow me to take AoPS classes because they cost like hundreds of dollars. I have to rely learning from online resources :/
Calculator User is in the same situation lol.
