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Use Pascal’s triangle to expand the binomial.

(d–5y)6


(2 points)

3 answers

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:

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

simple and short awnser and explanation

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