What is the coefficient of the xy^2 term in the expansion of (3x + y)^3? Use the binomial theorem.
In this case it is just as easy to simply expand it
= (3x)^3 + 3(3x)^2 y + 3(3x)(y^2) + y^3
so we want the second last term which is
9xy^2
To find the coefficient of the xy^2 term in the expansion of (3x + y)^3, we can use the binomial theorem.
The binomial theorem states that for any positive integer n, the expansion of (a + b)^n can be written as the sum of the terms of the form Ca,b * a^(n - b) * b^b, where Ca,b represents the binomial coefficient.
In this case, we have (3x + y)^3, so we need to find the term with xy^2.
The binomial coefficient Ca,b can be calculated using the formula:
Ca,b = n! / (b! * (n - b)!)
In this case, n = 3 and b = 2, so the binomial coefficient is:
C3,2 = 3! / (2! * (3 - 2)!) = 3! / (2! * 1!) = 3.
Now, let's substitute the values into the general term of the binomial expansion:
Term = Ca,b * a^(n - b) * b^b
Term = 3 * (3x)^(3 - 2) * (y^2)^2
Term = 3 * (3x) * y^4
Term = 9xy^4
Therefore, the coefficient of the xy^2 term in the expansion of (3x + y)^3 is 9.