Use the Binomial Theorem to find the fifth term in the expansion (2x+3y)^5.
5!/(5-k)^x^5-kyk
5!/(5-4)!4!^x^5-4y4
(5*4*3*2)/(4*3*2*1)
120/24
5xy^4 (answer)
You may not understand the 1st 2 lines I'm sorry just let me know how to make it better.
how about
C(5,4)(2x)^1(3y)^4
= 5*2*3^4*x*y^4
= 810xy^4
You have totally ignored the 2 and 3 in front of the x and y terms.
C(5,4) is read 5 choose 4 and has value
5!/(4!1!)
To find the fifth term in the expansion of (2x+3y)^5 using the Binomial Theorem, you need to determine the coefficient and the variables raised to their respective powers.
The Binomial Theorem states that the expansion of (a + b)^n can be written as the sum of terms of the form: C(n, k) * a^(n-k) * b^k, where C(n, k) represents the binomial coefficient "n choose k" and is calculated as n! / (k! * (n-k)!) and "!" denotes factorial.
In this case, (2x+3y)^5 can be expanded into the sum of terms:
C(5, 0) * (2x)^(5-0) * (3y)^0
C(5, 1) * (2x)^(5-1) * (3y)^1
C(5, 2) * (2x)^(5-2) * (3y)^2
C(5, 3) * (2x)^(5-3) * (3y)^3
C(5, 4) * (2x)^(5-4) * (3y)^4
C(5, 5) * (2x)^(5-5) * (3y)^5
Let's simplify each term:
Term 1: C(5, 0) * (2x)^5 * (3y)^0 = 1 * 2^5 * x^5 * 3^0 * y^0
Term 2: C(5, 1) * (2x)^4 * (3y)^1 = 5 * 2^4 * x^4 * 3^1 * y^1
Term 3: C(5, 2) * (2x)^3 * (3y)^2 = 10 * 2^3 * x^3 * 3^2 * y^2
Term 4: C(5, 3) * (2x)^2 * (3y)^3 = 10 * 2^2 * x^2 * 3^3 * y^3
Term 5: C(5, 4) * (2x)^1 * (3y)^4 = 5 * 2^1 * x^1 * 3^4 * y^4
Term 6: C(5, 5) * (2x)^0 * (3y)^5 = 1 * 2^0 * x^0 * 3^5 * y^5
The fifth term corresponds to Term 5: 5 * 2^1 * x^1 * 3^4 * y^4 = 10 * 2 * x * 81 * y^4 = 1620xy^4.
Therefore, the fifth term in the expansion of (2x+3y)^5 is 1620xy^4.