Question 975186
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I'm going to take a wild guess and presume you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{k=0}^4\,{4 \choose k}x^ky^{4-k}]


This sum has 5 terms, one for *[tex \LARGE k\ =\ 0], one for *[tex \LARGE k\ =\ 1], and so on.


The first term where *[tex \LARGE k\ =\ 0] is *[tex \LARGE {4 \choose 0}\,x^0y^{4-0}].  But *[tex \LARGE {n \choose 0}\ =\ 1\ \forall\ n\ \in\ \mathbb{Z}^+] and *[tex \LARGE x^0\ =\ 1\ \forall\ x\ \in\ \mathbb{R}], so the first term is simply *[tex \LARGE y^4]


The second term is where *[tex \LARGE k\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  {4 \choose 1}\ xy^3\ =\ 4xy^3]


because *[tex \LARGE {n \choose 1} = n] and *[tex \LARGE x^1\ =\ x]


Just follow the pattern for the other three terms.  You will have to calculate *[tex \LARGE {4 \choose 2}\ =\ \frac{4!}{2!(4-2)!}] for the third term but the other two will be easy because


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  {n \choose {n-1}}\ =\ {n \choose 1}\ =\ n]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ {n \choose n}\ =\ {n \choose 0}\ =\ 1]


for all positive integers *[tex \LARGE n]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \