%5E5.jpeg "(2x+3y)^5")
Expanding (2x + 3y)^5 Using the Binomial Theorem
The binomial theorem provides a formula for expanding expressions of the form (a + b)^n, where n is a positive integer. In this case, we want to expand (2x + 3y)^5.
Applying the Binomial Theorem
The binomial theorem states:
(a + b)^n = ∑ (n choose k) * a^(n-k) * b^k
Where:
- n choose k is the binomial coefficient, calculated as n! / (k! * (n-k)!).
- k ranges from 0 to n.
Let's apply this to our expression:
(2x + 3y)^5 = ∑ (5 choose k) * (2x)^(5-k) * (3y)^k
Expanding the Summation
We need to calculate the terms for each value of k from 0 to 5:
- k = 0: (5 choose 0) * (2x)^5 * (3y)^0 = 1 * 32x^5 * 1 = 32x^5
- k = 1: (5 choose 1) * (2x)^4 * (3y)^1 = 5 * 16x^4 * 3y = 240x^4y
- k = 2: (5 choose 2) * (2x)^3 * (3y)^2 = 10 * 8x^3 * 9y^2 = 720x^3y^2
- k = 3: (5 choose 3) * (2x)^2 * (3y)^3 = 10 * 4x^2 * 27y^3 = 1080x^2y^3
- k = 4: (5 choose 4) * (2x)^1 * (3y)^4 = 5 * 2x * 81y^4 = 810xy^4
- k = 5: (5 choose 5) * (2x)^0 * (3y)^5 = 1 * 1 * 243y^5 = 243y^5
Final Expanded Form
Combining all the terms, we get the complete expansion of (2x + 3y)^5:
(2x + 3y)^5 = 32x^5 + 240x^4y + 720x^3y^2 + 1080x^2y^3 + 810xy^4 + 243y^5
This demonstrates how the binomial theorem simplifies the expansion of complex expressions with exponents.