Answer :
To expand the binomial [tex]\((x - 5y)^5\)[/tex] using Pascal's triangle, follow these steps:
### Step 1: Identify the Coefficients from Pascal's Triangle
For a binomial expansion [tex]\((a + b)^n\)[/tex], the coefficients are the elements of the [tex]\(n\)[/tex]-th row of Pascal's triangle. For [tex]\(n = 5\)[/tex], the coefficients from Pascal's triangle are:
[tex]\[ 1, 5, 10, 10, 5, 1 \][/tex]
### Step 2: Set Up the Binomial Expansion
The general expansion formula is:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
Here, [tex]\(a = x\)[/tex] and [tex]\(b = -5y\)[/tex].
### Step 3: Apply the Formula
Substitute [tex]\(x\)[/tex] for [tex]\(a\)[/tex] and [tex]\(-5y\)[/tex] for [tex]\(b\)[/tex], and use the coefficients from Step 1:
1. [tex]\( \binom{5}{0} x^5 (-5y)^0 = 1 \cdot x^5 \cdot 1 = x^5 \)[/tex]
2. [tex]\( \binom{5}{1} x^4 (-5y)^1 = 5 \cdot x^4 \cdot (-5y) = -25x^4 y \)[/tex]
3. [tex]\( \binom{5}{2} x^3 (-5y)^2 = 10 \cdot x^3 \cdot 25y^2 = 250x^3 y^2 \)[/tex]
4. [tex]\( \binom{5}{3} x^2 (-5y)^3 = 10 \cdot x^2 \cdot (-125y^3) = -1250x^2 y^3 \)[/tex]
5. [tex]\( \binom{5}{4} x^1 (-5y)^4 = 5 \cdot x \cdot 625y^4 = 3125x y^4 \)[/tex]
6. [tex]\( \binom{5}{5} x^0 (-5y)^5 = 1 \cdot 1 \cdot (-3125y^5) = -3125y^5 \)[/tex]
### Step 4: Combine the Terms
Combining all the terms gives the expanded form:
[tex]\[ x^5 - 25x^4y + 250x^3y^2 - 1250x^2y^3 + 3125xy^4 - 3125y^5 \][/tex]
This expanded form corresponds to the choice:
[tex]\[ x^5 - 25x^4 y + 250x^3 y^2 - 1250x^2 y^3 + 3125x y^4 - 3125y^5 \][/tex]
Therefore, the correct expanded form of [tex]\((x - 5y)^5\)[/tex] is:
[tex]\[ x^5 - 25x^4 y + 250x^3 y^2 - 1250x^2 y^3 + 3125x y^4 - 3125y^5 \][/tex]
### Step 1: Identify the Coefficients from Pascal's Triangle
For a binomial expansion [tex]\((a + b)^n\)[/tex], the coefficients are the elements of the [tex]\(n\)[/tex]-th row of Pascal's triangle. For [tex]\(n = 5\)[/tex], the coefficients from Pascal's triangle are:
[tex]\[ 1, 5, 10, 10, 5, 1 \][/tex]
### Step 2: Set Up the Binomial Expansion
The general expansion formula is:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
Here, [tex]\(a = x\)[/tex] and [tex]\(b = -5y\)[/tex].
### Step 3: Apply the Formula
Substitute [tex]\(x\)[/tex] for [tex]\(a\)[/tex] and [tex]\(-5y\)[/tex] for [tex]\(b\)[/tex], and use the coefficients from Step 1:
1. [tex]\( \binom{5}{0} x^5 (-5y)^0 = 1 \cdot x^5 \cdot 1 = x^5 \)[/tex]
2. [tex]\( \binom{5}{1} x^4 (-5y)^1 = 5 \cdot x^4 \cdot (-5y) = -25x^4 y \)[/tex]
3. [tex]\( \binom{5}{2} x^3 (-5y)^2 = 10 \cdot x^3 \cdot 25y^2 = 250x^3 y^2 \)[/tex]
4. [tex]\( \binom{5}{3} x^2 (-5y)^3 = 10 \cdot x^2 \cdot (-125y^3) = -1250x^2 y^3 \)[/tex]
5. [tex]\( \binom{5}{4} x^1 (-5y)^4 = 5 \cdot x \cdot 625y^4 = 3125x y^4 \)[/tex]
6. [tex]\( \binom{5}{5} x^0 (-5y)^5 = 1 \cdot 1 \cdot (-3125y^5) = -3125y^5 \)[/tex]
### Step 4: Combine the Terms
Combining all the terms gives the expanded form:
[tex]\[ x^5 - 25x^4y + 250x^3y^2 - 1250x^2y^3 + 3125xy^4 - 3125y^5 \][/tex]
This expanded form corresponds to the choice:
[tex]\[ x^5 - 25x^4 y + 250x^3 y^2 - 1250x^2 y^3 + 3125x y^4 - 3125y^5 \][/tex]
Therefore, the correct expanded form of [tex]\((x - 5y)^5\)[/tex] is:
[tex]\[ x^5 - 25x^4 y + 250x^3 y^2 - 1250x^2 y^3 + 3125x y^4 - 3125y^5 \][/tex]
