Question 213890
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x + y\right)^n = \sum_{k=0}^n\,\left(n \cr k\right)\,x^{n-k}y^k]


where


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(n \cr k\right) = \frac{n!}{k!\left(n-k\right)!}]


For your case,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x + y\right)^7]


*[tex \Large n = 7] and *[tex \Large k] runs from 0 to 7, so the coefficient of the first term of your expansion is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(7 \cr 0\right) = \frac{7!}{0!\left(7-0\right)!}] (remember 0! = 1) *[tex \LARGE = \frac{7!}{7!} = 1]


The rest of the term is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^{7-0}y^0 = x^7]


Therefore your entire 1st term is: *[tex \LARGE x^7]


Now let's work on the 2nd term, coefficient first:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(7 \cr 1\right) = \frac{7!}{1!\left(7-1\right)!}\ =\ \frac{7!}{6!} = 7]


Then the variables:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^{7-1}y^1 = x^6y^1 = x^6y]


Entire 2nd term:  *[tex \LARGE 7x^6y]


Third term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(7 \cr 2\right) = \frac{7!}{2!\left(7-2\right)!}\ =\ \frac{7!}{2 \cdot 5!} = \frac{7\cdot6}{2} = 21]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^{7-2}y^2 = x^5y^2]


Hence, *[tex \LARGE 21x^5y^2]


Just follow the pattern to do the rest of the terms and show your answer as the sum of all the terms.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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