Find the binomial expansion of (2x+y)^4
To find the binomial expansion of (2x + y)^4, we can use the Binomial Theorem.
The Binomial Theorem states that for any positive integer n:
(x + y)^n = C(n, 0) * x^n * y^0 + C(n, 1) * x^(n-1) * y^1 + C(n, 2) * x^(n-2) * y^2 + ... + C(n, n-1) * x^1 * y^(n-1) + C(n, n) * x^0 * y^n
where C(n, k) represents the binomial coefficient, defined as C(n, k) = n! / (k!(n-k)!). Therefore, we can find the binomial expansion of (2x + y)^4 as follows:
(2x + y)^4 = C(4, 0) * (2x)^4 * y^0 + C(4, 1) * (2x)^3 * y^1 + C(4, 2) * (2x)^2 * y^2 + C(4, 3) * (2x)^1 * y^3 + C(4, 4) * (2x)^0 * y^4
Simplifying each term, we get:
(2x + y)^4 = 1 * 16x^4 * 1 + 4 * 8x^3 * y + 6 * 4x^2 * y^2 + 4 * 2x * y^3 + 1 * 1 * y^4
(2x + y)^4 = 16x^4 + 32x^3y + 24x^2y^2 + 8xy^3 + y^4
Therefore, the binomial expansion of (2x + y)^4 is:
16x^4 + 32x^3y + 24x^2y^2 + 8xy^3 + y^4
To find the binomial expansion of (2x + y)^4, you 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:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
Where C(n, k) denotes combinations, which can be calculated using n! / (k!(n-k)!).
For (2x + y)^4, the coefficients and powers of a are as follows:
C(4, 0) * (2x)^4 * y^0 = 16x^4
C(4, 1) * (2x)^3 * y^1 = 96x^3y
C(4, 2) * (2x)^2 * y^2 = 144x^2y^2
C(4, 3) * (2x)^1 * y^3 = 96xy^3
C(4, 4) * (2x)^0 * y^4 = y^4
Therefore, the binomial expansion of (2x + y)^4 is:
16x^4 + 96x^3y + 144x^2y^2 + 96xy^3 + y^4
To find the binomial expansion of (2x+y)^4, we can use the binomial theorem. The binomial theorem states that for any positive integer n:
(x + y)^n = C(n,0) * x^n * y^0 + C(n,1) * x^(n-1) * y^1 + C(n,2) * x^(n-2) * y^2 + ... + C(n, n-1) * x^1 * y^(n-1) + C(n,n) * x^0 * y^n
Where C(n,k) = n! / (k! * (n-k)!)
Let's apply the binomial theorem to (2x+y)^4:
(2x + y)^4 = C(4,0) * (2x)^4 * y^0 + C(4,1) * (2x)^3 * y^1 + C(4,2) * (2x)^2 * y^2 + C(4,3) * (2x)^1 * y^3 + C(4,4) * (2x)^0 * y^4
Now, let's calculate the binomial coefficients C(n,k):
C(4,0) = 4! / (0! * (4-0)!) = 1
C(4,1) = 4! / (1! * (4-1)!) = 4
C(4,2) = 4! / (2! * (4-2)!) = 6
C(4,3) = 4! / (3! * (4-3)!) = 4
C(4,4) = 4! / (4! * (4-4)!) = 1
Now, we substitute the values into the expansion:
(2x + y)^4 = 1 * (2x)^4 * y^0 + 4 * (2x)^3 * y^1 + 6 * (2x)^2 * y^2 + 4 * (2x)^1 * y^3 + 1 * (2x)^0 * y^4
Simplifying each term:
(2x)^4 * y^0 = 16x^4
4 * (2x)^3 * y^1 = 32x^3y
6 * (2x)^2 * y^2 = 24x^2y^2
4 * (2x)^1 * y^3 = 8xy^3
(2x)^0 * y^4 = y^4
Therefore, the binomial expansion of (2x+y)^4 is:
16x^4 + 32x^3y + 24x^2y^2 + 8xy^3 + y^4