(x+y)⁵ expression patulong naman po step by step
Answer (1)
Answer:Therefore, the expanded expression is x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5.Step-by-step explanation:To expand the expression (x+y)^5, we can use the binomial theorem. The binomial theorem states that for any non-negative integer n: (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k where \binom{n}{k} = \frac{n!}{k!(n-k)!} is the binomial coefficient, also written as _nC_k or C(n,k). Let's apply this to (x+y)^5: Step 1: Identify n, a, and b In our case, n=5, a=x, and b=y. Step 2: Apply the Binomial Theorem (x+y)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} y^k Step 3: Expand the summation This means we'll have terms for k = 0, 1, 2, 3, 4, and 5. Let's calculate each term: - k = 0: \binom{5}{0} x^{5-0} y^0 = 1 \cdot x^5 \cdot 1 = x^5- k = 1: \binom{5}{1} x^{5-1} y^1 = 5 \cdot x^4 \cdot y = 5x^4y- k = 2: \binom{5}{2} x^{5-2} y^2 = 10 \cdot x^3 \cdot y^2 = 10x^3y^2- k = 3: \binom{5}{3} x^{5-3} y^3 = 10 \cdot x^2 \cdot y^3 = 10x^2y^3- k = 4: \binom{5}{4} x^{5-4} y^4 = 5 \cdot x^1 \cdot y^4 = 5xy^4- k = 5: \binom{5}{5} x^{5-5} y^5 = 1 \cdot 1 \cdot y^5 = y^5 Step 4: Combine the terms Adding all the terms together, we get the final expansion: (x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5
