+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.
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