Expand (2x+3y)^4
Using the binomial theorem
The binomial theorem states that for any real numbers a and b, and any positive integer n,
(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 the binomial coefficient, which is calculated as C(n,k) = n! / (k! * (n-k)!)
In this case, we have a = 2x, b = 3y, and n = 4.
So, applying the binomial theorem, we have:
(2x + 3y)^4 = C(4,0)(2x)^4 * (3y)^0 + C(4,1)(2x)^3 * (3y)^1 + C(4,2)(2x)^2 * (3y)^2 + C(4,3)(2x)^1 * (3y)^3 + C(4,4)(2x)^0 * (3y)^4
Simplifying each term, we get:
(2x + 3y)^4 = 1(2x)^4 * 1 + 4(2x)^3 * (3y)^1 + 6(2x)^2 * (3y)^2 + 4(2x)^1 * (3y)^3 + 1(2x)^0 * (3y)^4
Expanding further, we get:
(2x + 3y)^4 = 16x^4 + 96x^3y + 216x^2y^2 + 216xy^3 + 81y^4
To expand (2x+3y)^4 using the binomial theorem, we can use the formula:
(a + b)^n = nC0 * a^n * b^0 + nC1 * a^(n-1) * b^1 + nC2 * a^(n-2) * b^2 + ... + nCr * a^(n-r) * b^r + ... + nCn * a^0 * b^n
Where n is the power of the binomial, a represents the first term (2x in this case), b represents the second term (3y in this case), and nCr represents the combination formula n! / (r! * (n-r)!).
Using this formula, let's expand (2x+3y)^4 step by step:
Step 1: Calculate the values of nCr:
nC0 = 4! / (0! * (4-0)!) = 1
nC1 = 4! / (1! * (4-1)!) = 4
nC2 = 4! / (2! * (4-2)!) = 6
nC3 = 4! / (3! * (4-3)!) = 4
nC4 = 4! / (4! * (4-4)!) = 1
Step 2: Substitute the values into the formula:
(2x + 3y)^4 = 1 * (2x)^4 * (3y)^0 + 4 * (2x)^3 * (3y)^1 + 6 * (2x)^2 * (3y)^2 + 4 * (2x)^1 * (3y)^3 + 1 * (2x)^0 * (3y)^4
Step 3: Simplify the exponents:
(2x + 3y)^4 = 1 * 2^4 * x^4 * 1 + 4 * 2^3 * x^3 * 3y + 6 * 2^2 * x^2 * (3y)^2 + 4 * 2^1 * x * (3y)^3 + 1 * 1 * (3y)^4
Step 4: Evaluate the powers:
(2x + 3y)^4 = 16x^4 + 96x^3y + 216x^2y^2 + 216xy^3 + 81y^4
Therefore, the expanded form of (2x+3y)^4 using the binomial theorem is 16x^4 + 96x^3y + 216x^2y^2 + 216xy^3 + 81y^4.