Question

question 9 1 pts use pascals triangle to expand the binomial ((x + 1)^5). ( circ x^5 - 5x^4 - 10x^3 - 10x^2 - 5x - 1 ) ( circ x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1 ) ( circ x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1 ) ( circ x^5 - 5x^4 - 10x^3 - 10x^2 + 5x - 1 )

Explanation:

Step1: Recall Pascal's Triangle for \( n = 5 \)

Pascal's Triangle row for \( n = 5 \) (where the first row is \( n = 0 \)) is \( 1, 5, 10, 10, 5, 1 \).

Step2: Apply Binomial Expansion Formula

The binomial expansion of \( (a + b)^n \) is \( \sum_{k = 0}^{n} \binom{n}{k} a^{n - k} b^{k} \). For \( (x + 1)^5 \), \( a = x \), \( b = 1 \), and \( n = 5 \). Using the coefficients from Pascal's Triangle:

• When \( k = 0 \): \( \binom{5}{0} x^{5 - 0} 1^{0} = 1 \cdot x^5 \cdot 1 = x^5 \)
• When \( k = 1 \): \( \binom{5}{1} x^{5 - 1} 1^{1} = 5 \cdot x^4 \cdot 1 = 5x^4 \)
• When \( k = 2 \): \( \binom{5}{2} x^{5 - 2} 1^{2} = 10 \cdot x^3 \cdot 1 = 10x^3 \)
• When \( k = 3 \): \( \binom{5}{3} x^{5 - 3} 1^{3} = 10 \cdot x^2 \cdot 1 = 10x^2 \)
• When \( k = 4 \): \( \binom{5}{4} x^{5 - 4} 1^{4} = 5 \cdot x^1 \cdot 1 = 5x \)
• When \( k = 5 \): \( \binom{5}{5} x^{5 - 5} 1^{5} = 1 \cdot x^0 \cdot 1 = 1 \)

Step3: Combine Terms

Adding these terms together: \( x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1 \).

Answer:

\( x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1 \) (corresponding to the option \( x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1 \))