Answer :
Sure! Let's expand the binomial [tex]\((x - 5y)^5\)[/tex] using Pascal's Triangle.
Here's a step-by-step process:
1. Identify Pascal's Triangle Coefficients:
For expanding [tex]\((a + b)^n\)[/tex], we use the [tex]\(n\)[/tex]th row of Pascal's Triangle. For [tex]\(n = 5\)[/tex], the coefficients are:
[tex]\[
\binom{5}{0}, \binom{5}{1}, \binom{5}{2}, \binom{5}{3}, \binom{5}{4}, \binom{5}{5}
\][/tex]
These coefficients are [tex]\(1, 5, 10, 10, 5, 1\)[/tex].
2. Write the General Form of the Expansion:
The binomial expansion of [tex]\((x - 5y)^5\)[/tex] is:
[tex]\[
\sum_{k=0}^{5} \binom{5}{k} (x)^{5-k} (-5y)^k
\][/tex]
3. Calculate Each Term:
- For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{5}{0} x^{5-0} (-5y)^0 = 1 \cdot x^5 \cdot 1 = x^5
\][/tex]
- For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{5}{1} x^{5-1} (-5y)^1 = 5 \cdot x^4 \cdot (-5y) = -25x^4y
\][/tex]
- For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{5}{2} x^{5-2} (-5y)^2 = 10 \cdot x^3 \cdot (25y^2) = 250x^3 y^2
\][/tex]
- For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{5}{3} x^{5-3} (-5y)^3 = 10 \cdot x^2 \cdot (-125y^3) = -1250x^2 y^3
\][/tex]
- For [tex]\(k = 4\)[/tex]:
[tex]\[
\binom{5}{4} x^{5-4} (-5y)^4 = 5 \cdot x^1 \cdot 625y^4 = 3125x y^4
\][/tex]
- For [tex]\(k = 5\)[/tex]:
[tex]\[
\binom{5}{5} x^{5-5} (-5y)^5 = 1 \cdot 1 \cdot (-3125y^5) = -3125y^5
\][/tex]
4. Combine All Terms Together:
[tex]\[
x^5 - 25x^4y + 250x^3y^2 - 1250x^2y^3 + 3125xy^4 - 3125y^5
\][/tex]
So, 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]
This matches option [tex]\( \mathbf{x^5 - 25 x^4 y + 250 x^3 y^2 - 1250 x^2 y^3 + 3125 x y^4 - 3125 y^5} \)[/tex].
Here's a step-by-step process:
1. Identify Pascal's Triangle Coefficients:
For expanding [tex]\((a + b)^n\)[/tex], we use the [tex]\(n\)[/tex]th row of Pascal's Triangle. For [tex]\(n = 5\)[/tex], the coefficients are:
[tex]\[
\binom{5}{0}, \binom{5}{1}, \binom{5}{2}, \binom{5}{3}, \binom{5}{4}, \binom{5}{5}
\][/tex]
These coefficients are [tex]\(1, 5, 10, 10, 5, 1\)[/tex].
2. Write the General Form of the Expansion:
The binomial expansion of [tex]\((x - 5y)^5\)[/tex] is:
[tex]\[
\sum_{k=0}^{5} \binom{5}{k} (x)^{5-k} (-5y)^k
\][/tex]
3. Calculate Each Term:
- For [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{5}{0} x^{5-0} (-5y)^0 = 1 \cdot x^5 \cdot 1 = x^5
\][/tex]
- For [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{5}{1} x^{5-1} (-5y)^1 = 5 \cdot x^4 \cdot (-5y) = -25x^4y
\][/tex]
- For [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{5}{2} x^{5-2} (-5y)^2 = 10 \cdot x^3 \cdot (25y^2) = 250x^3 y^2
\][/tex]
- For [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{5}{3} x^{5-3} (-5y)^3 = 10 \cdot x^2 \cdot (-125y^3) = -1250x^2 y^3
\][/tex]
- For [tex]\(k = 4\)[/tex]:
[tex]\[
\binom{5}{4} x^{5-4} (-5y)^4 = 5 \cdot x^1 \cdot 625y^4 = 3125x y^4
\][/tex]
- For [tex]\(k = 5\)[/tex]:
[tex]\[
\binom{5}{5} x^{5-5} (-5y)^5 = 1 \cdot 1 \cdot (-3125y^5) = -3125y^5
\][/tex]
4. Combine All Terms Together:
[tex]\[
x^5 - 25x^4y + 250x^3y^2 - 1250x^2y^3 + 3125xy^4 - 3125y^5
\][/tex]
So, 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]
This matches option [tex]\( \mathbf{x^5 - 25 x^4 y + 250 x^3 y^2 - 1250 x^2 y^3 + 3125 x y^4 - 3125 y^5} \)[/tex].
