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Answer :
To expand the binomial [tex]\((x - 5y)^5\)[/tex] using Pascal's triangle, we can follow these steps:
### Step 1: Identify the Coefficients
Using Pascal's triangle for the 5th power, we find the coefficients: 1, 5, 10, 10, 5, 1.
### Step 2: Apply the Binomial Theorem
The binomial theorem states that [tex]\((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)[/tex]. Here, [tex]\(a = x\)[/tex], [tex]\(b = -5y\)[/tex], and [tex]\(n = 5\)[/tex].
### Step 3: Calculate Each Term
1. First Term:
[tex]\[
\binom{5}{0} \cdot x^5 \cdot (-5y)^0 = 1 \cdot x^5 \cdot 1 = x^5
\][/tex]
2. Second Term:
[tex]\[
\binom{5}{1} \cdot x^4 \cdot (-5y)^1 = 5 \cdot x^4 \cdot (-5y) = -25x^4y
\][/tex]
3. Third Term:
[tex]\[
\binom{5}{2} \cdot x^3 \cdot (-5y)^2 = 10 \cdot x^3 \cdot 25y^2 = 250x^3y^2
\][/tex]
4. Fourth Term:
[tex]\[
\binom{5}{3} \cdot x^2 \cdot (-5y)^3 = 10 \cdot x^2 \cdot (-125y^3) = -1250x^2y^3
\][/tex]
5. Fifth Term:
[tex]\[
\binom{5}{4} \cdot x^1 \cdot (-5y)^4 = 5 \cdot x \cdot 625y^4 = 3125xy^4
\][/tex]
6. Sixth Term:
[tex]\[
\binom{5}{5} \cdot x^0 \cdot (-5y)^5 = 1 \cdot 1 \cdot (-3125y^5) = -3125y^5
\][/tex]
### Step 4: Combine All Terms
Putting all these components together, the expanded form of [tex]\((x - 5y)^5\)[/tex] is:
[tex]\[
x^5 - 25x^4y + 250x^3y^2 - 1250x^2y^3 + 3125xy^4 - 3125y^5
\][/tex]
That's the expanded form of the binomial [tex]\((x - 5y)^5\)[/tex] using Pascal's triangle.
### Step 1: Identify the Coefficients
Using Pascal's triangle for the 5th power, we find the coefficients: 1, 5, 10, 10, 5, 1.
### Step 2: Apply the Binomial Theorem
The binomial theorem states that [tex]\((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)[/tex]. Here, [tex]\(a = x\)[/tex], [tex]\(b = -5y\)[/tex], and [tex]\(n = 5\)[/tex].
### Step 3: Calculate Each Term
1. First Term:
[tex]\[
\binom{5}{0} \cdot x^5 \cdot (-5y)^0 = 1 \cdot x^5 \cdot 1 = x^5
\][/tex]
2. Second Term:
[tex]\[
\binom{5}{1} \cdot x^4 \cdot (-5y)^1 = 5 \cdot x^4 \cdot (-5y) = -25x^4y
\][/tex]
3. Third Term:
[tex]\[
\binom{5}{2} \cdot x^3 \cdot (-5y)^2 = 10 \cdot x^3 \cdot 25y^2 = 250x^3y^2
\][/tex]
4. Fourth Term:
[tex]\[
\binom{5}{3} \cdot x^2 \cdot (-5y)^3 = 10 \cdot x^2 \cdot (-125y^3) = -1250x^2y^3
\][/tex]
5. Fifth Term:
[tex]\[
\binom{5}{4} \cdot x^1 \cdot (-5y)^4 = 5 \cdot x \cdot 625y^4 = 3125xy^4
\][/tex]
6. Sixth Term:
[tex]\[
\binom{5}{5} \cdot x^0 \cdot (-5y)^5 = 1 \cdot 1 \cdot (-3125y^5) = -3125y^5
\][/tex]
### Step 4: Combine All Terms
Putting all these components together, the expanded form of [tex]\((x - 5y)^5\)[/tex] is:
[tex]\[
x^5 - 25x^4y + 250x^3y^2 - 1250x^2y^3 + 3125xy^4 - 3125y^5
\][/tex]
That's the expanded form of the binomial [tex]\((x - 5y)^5\)[/tex] using Pascal's triangle.
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