SOLUTION: Consider the expansion of (x+y)^n. 1. How many terms does the expression contain? 2. What is the exponent of x in the first term? 3. What is the exponent of y in the first te

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Consider the expansion of (x+y)^n. 1. How many terms does the expression contain? 2. What is the exponent of x in the first term? 3. What is the exponent of y in the first te      Log On


   

Question 1002501: Consider the expansion of (x+y)^n.
1. How many terms does the expression contain?
2. What is the exponent of x in the first term?
3. What is the exponent of y in the first term?
4. What is the sum of the exponents in any term of the expansion ?
Show work please, thanks
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!

is a formula/theorem that is usually proven in class, and then the proof is often forgotten.
1. The expression contain n%2B1 terms,
which is easy to see/count in the formula above,
because you see 1 first x%5En term without a y , plus
n terms showing y with n different numbers as exponents,
starting with the invisible exponent 1 in the term n%2Ax%5E%28n-1%29y=n%2Ax%5E%28n-1%29y%5E1 ,
and going all the way to exponent n in the y%5En last term.
2. The exponent of x in the first term is n because the term that we write first is x%5En .
There is no compelling reason to write the terms in that order,
but it is customary, and having a certain order in mind helps keep track of all those terms.
3. Since the first term is x%5En=x%5En%2A1=x%5En%2Ay%5E0 , we can say that the exponent of y in the first term is 0 .
4. The sum of the exponents in any term of the expansion is n .
That is true for the first term, x%5En=x%5En%2Ay%5E0 , and is also true for all of the other terms.

EXPLANATION OF THE FORMULA (in case you care):
The formula comes from the fact that
%28x%2By%29%5En=%28x%2By%29%2A%28x%2By%29%2A%22...%22%2A%28x%2By%29%2A%28x%2By%29 with n factors.
Before simplifying, the product of those n factors would have 2%5En products made by choosing one of the variables (x or y) from each of the %28x%2By%29 factors.
Because each of those 2%5En products has n factors, the degree of each product (meaning the sum of the exponents of x and y) is n .
Choosing the x from each of the n %28x%2By%29 factors, we would get x%5En . The only one way to get the product x%5En is to choose the x term from all of the n %28x%2By%29 factors,
so you get that product only once.
The same can be said of y%5En .
Because we like to put x's before y's, and the %28x%2By%29%5En already had the x before the y,
x%5En is written as the first term, and y%5En is written as the last term.
We get other products multiple times
If you choose the y from one of the %28x%2By%29 factors and the x from the others, you get products like
y%2Ax%2Ax%2Ax%2A%22...%22x%2Ax , x%2Ay%2Ax%2Ax%2A%22...%22x%2Ax that can be written as x%5E%28n-1%29%2Ay .
Of course there are n different ways to do that, and after simplifying all those n products would be accounted for in the term n%2Ax%5E%28n-1%29%2Ay .
There are n%28n-1%29%2F2 ways to choose 2 y's and n-2 x's,
and there are also n%28n-1%29%2F2 ways to choose 2 x's and n-2 y's,
and that explains the coefficients in the terms %28n%28n-1%29%2F2%29x%5E%28n-2%29y%5E2 and %28n%28n-1%29%2F2%29x%5E2y%5E%28n-2%29 .