Question

Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.

Use Pascal’s triangle to expand the binomial.

(d–5y)6


(2 points)

Answer 1

simple and short awnser and explanation

Answer 2

To expand the binomial \((d - 5y)^6\) using Pascal's triangle, we first need to find the coefficients from the 6th row of Pascal's triangle.

### Step 1: Identify the coefficients from Pascal's Triangle
The rows of Pascal's triangle are as follows:

- Row 0: \( \binom{0}{0} = 1 \)
- Row 1: \( \binom{1}{0} = 1, \binom{1}{1} = 1 \)
- Row 2: \( \binom{2}{0} = 1, \binom{2}{1} = 2, \binom{2}{2} = 1 \)
- Row 3: \( \binom{3}{0} = 1, \binom{3}{1} = 3, \binom{3}{2} = 3, \binom{3}{3} = 1 \)
- Row 4: \( \binom{4}{0} = 1, \binom{4}{1} = 4, \binom{4}{2} = 6, \binom{4}{3} = 4, \binom{4}{4} = 1 \)
- Row 5: \( \binom{5}{0} = 1, \binom{5}{1} = 5, \binom{5}{2} = 10, \binom{5}{3} = 10, \binom{5}{4} = 5, \binom{5}{5} = 1 \)
- Row 6: \( \binom{6}{0} = 1, \binom{6}{1} = 6, \binom{6}{2} = 15, \binom{6}{3} = 20, \binom{6}{4} = 15, \binom{6}{5} = 6, \binom{6}{6} = 1 \)

So, the coefficients for the expansion of \((d - 5y)^6\) will be:
1, 6, 15, 20, 15, 6, 1.

### Step 2: Write the expansion
Using the Binomial Theorem, we can expand \((a + b)^n\) as:

\[
\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

In our case, \(a = d\), \(b = -5y\), and \(n = 6\).

### Step 3: Apply the Binomial Theorem
Thus, we can write:

\[
(d - 5y)^6 = \sum_{k=0}^{6} \binom{6}{k} d^{6-k} (-5y)^k
\]

Now let's calculate each term for \(k = 0\) to \(k = 6\):

- For \(k = 0\):
\[
\binom{6}{0} d^{6} (-5y)^0 = 1 \cdot d^6 \cdot 1 = d^6
\]

- For \(k = 1\):
\[
\binom{6}{1} d^{5} (-5y)^1 = 6 \cdot d^5 \cdot (-5y) = -30d^5y
\]

- For \(k = 2\):
\[
\binom{6}{2} d^{4} (-5y)^2 = 15 \cdot d^4 \cdot 25y^2 = 375d^4y^2
\]

- For \(k = 3\):
\[
\binom{6}{3} d^{3} (-5y)^3 = 20 \cdot d^3 \cdot (-125y^3) = -2500d^3y^3
\]

- For \(k = 4\):
\[
\binom{6}{4} d^{2} (-5y)^4 = 15 \cdot d^2 \cdot 625y^4 = 9375d^2y^4
\]

- For \(k = 5\):
\[
\binom{6}{5} d^{1} (-5y)^5 = 6 \cdot d \cdot (-3125y^5) = -18750dy^5
\]

- For \(k = 6\):
\[
\binom{6}{6} d^{0} (-5y)^6 = 1 \cdot 1 \cdot 15625y^6 = 15625y^6
\]

### Step 4: Combine all terms
Now combining all these terms gives us:

\[
(d - 5y)^6 = d^6 - 30d^5y + 375d^4y^2 - 2500d^3y^3 + 9375d^2y^4 - 18750dy^5 + 15625y^6
\]

### Final Answer:
The expansion of \((d - 5y)^6\) is:

\[
\boxed{d^6 - 30d^5y + 375d^4y^2 - 2500d^3y^3 + 9375d^2y^4 - 18750dy^5 + 15625y^6}
\]

Answer 3

To expand \((d - 5y)^6\) using Pascal's triangle, follow these steps:

### Step 1: Coefficients from Pascal's Triangle (Row 6)
The coefficients for the expansion are:
1, 6, 15, 20, 15, 6, 1.

### Step 2: Binomial Expansion
Using the binomial expansion formula:
\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]
where \(a = d\), \(b = -5y\), and \(n = 6\), the expanded form is:
\[
= \binom{6}{0} d^6 (-5y)^0 + \binom{6}{1} d^5 (-5y)^1 + \binom{6}{2} d^4 (-5y)^2 + \ldots + \binom{6}{6} d^0 (-5y)^6
\]

### Step 3: Combine Terms
Calculating each term gives:
\[
= d^6 - 30d^5y + 375d^4y^2 - 2500d^3y^3 + 9375d^2y^4 - 18750dy^5 + 15625y^6
\]

### Final Answer:
The expanded form is:
\[
\boxed{d^6 - 30d^5y + 375d^4y^2 - 2500d^3y^3 + 9375d^2y^4 - 18750dy^5 + 15625y^6}
\]