Question

question
use pascals triangle to expand ((5y - 1)^5). express your answer in simplest form.
answer attempt 2 out of 2

Explanation:

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

The 5th row of Pascal's Triangle (starting from row 0) is \(1, 5, 10, 10, 5, 1\). The binomial expansion formula is \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}\) are the coefficients from Pascal's Triangle. Here, \(a = 5y\), \(b=- 1\), and \(n = 5\).

Step2: Expand using the coefficients

\[

$$\begin{align*} (5y-1)^{5}&=1\times(5y)^{5}\times(-1)^{0}+5\times(5y)^{4}\times(-1)^{1}+10\times(5y)^{3}\times(-1)^{2}+10\times(5y)^{2}\times(-1)^{3}+5\times(5y)^{1}\times(-1)^{4}+1\times(5y)^{0}\times(-1)^{5}\\ &=(5y)^{5}+5\times(5y)^{4}\times(-1)+10\times(5y)^{3}\times1 + 10\times(5y)^{2}\times(-1)+5\times(5y)\times1+1\times1\times(-1) \end{align*}$$

\]

Step3: Simplify each term

• For the first term: \((5y)^{5}=5^{5}y^{5}=3125y^{5}\)
• Second term: \(5\times(5y)^{4}\times(-1)=5\times625y^{4}\times(-1)=- 3125y^{4}\)
• Third term: \(10\times(5y)^{3}\times1=10\times125y^{3}=1250y^{3}\)
• Fourth term: \(10\times(5y)^{2}\times(-1)=10\times25y^{2}\times(-1)=-250y^{2}\)
• Fifth term: \(5\times(5y)\times1 = 25y\)
• Sixth term: \(1\times1\times(-1)=-1\)

Step4: Combine all terms

\[
(5y - 1)^{5}=3125y^{5}-3125y^{4}+1250y^{3}-250y^{2}+25y - 1
\]

Answer:

\(3125y^{5}-3125y^{4}+1250y^{3}-250y^{2}+25y - 1\)