Question 1002539
# 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.