Question 1002539: Consider the expansion of (x+y)^n. thanks!
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 jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! # 1
If n = 1, there are 2 terms since (x+y)^n = (x+y)^1 = x+y
If n = 2, there are 3 terms since (x+y)^n = (x+y)^2 = x^2+2xy+y^2
If n = 3, there are 4 terms since (x+y)^n = (x+y)^3 = x^3+3x^2y+3xy^2+y^3
and so on
The pattern continues. For any positive integer n, there are n+1 terms in (x+y)^n
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# 2
If n = 1, then (x+y)^n = (x+y)^1 = x+y
First term is x = x^1
If n = 2, then (x+y)^n = (x+y)^2 = x^2+2xy+y^2
First term is x^2
If n = 3, then (x+y)^n = (x+y)^3 = x^3+3x^2y+3xy^2+y^3
First term is x^3
I'm sure you see the pattern here as well
The first term of (x+y)^n is x^n. There are NO y terms in the first term
The exponent for the x in the first term is n
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# 3
See problem #2 above. The first term is x^n which is really the same as x^n*y^0. So the exponent for y in the first term is 0
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# 4
The sum of exponents of any monomial term is always n
Take a look at (x+y)^3 = x^3+3x^2y+3xy^2+y^3, where n = 3 in this case
Then focus on a term like 3x^2y. This term is the same as 3x^2*y^1. The exponents are 2 and 1 which add to 2+1 = 3 which is the value of n.
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