5 a Write down the first 5 rows of Pascal's triangle. 6 What is the sum of the numbers in: 1 row 1 if row 2 It row 3 tv row 4 V row 57 c Copy and completc: The sum of the numbers in rown of Pascal's triengle is \( \qquad \) d Show that \( (1+x)^{n}=\binom{n}{0}+\binom{n}{1} x+\binom{n}{2} x^{2}+\ldots+\binom{n}{n-1} x^{n-1}+\binom{n}{n} x^{n} \) Hence deduce that \( \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\ldots .+\binom{n}{n-1}+\binom{n}{n}=2^{n} \)

Let's break down the problem step by step. ### Part 5a: First 5 Rows of Pascal's Triangle Pascal's triangle is constructed such that each number is the sum of the two numbers directly above it. The first five rows are: ``` Row 0: 1 Row 1: 1 1 Row 2: 1 2 1 Row 3: 1 3 3 1 Row 4: 1 4 6 4 1 ``` ### Part 6: Sum of the Numbers in Each Row 1. **Row 0**: The sum is \( 1 \). 2. **Row 1**: The sum is \( 1 + 1 = 2 \). 3. **Row 2**: The sum is \( 1 + 2 + 1 = 4 \). 4. **Row 3**: The sum is \( 1 + 3 + 3 + 1 = 8 \). 5. **Row 4**: The sum is \( 1 + 4 + 6 + 4 + 1 = 16 \). 6. **Row 57**: The sum can be calculated using the formula \( 2^n \) where \( n \) is the row number. Thus, the sum for row 57 is \( 2^{57} \). ### Part 6c: Completing the Statement The sum of the numbers in row \( n \) of Pascal's triangle is \( 2^n \). ### Part 6d: Showing the Binomial Expansion To show that \( (1+x)^{n} = \binom{n}{0} + \binom{n}{1} x + \binom{n}{2} x^{2} + \ldots + \binom{n}{n} x^{n} \), we can use the Binomial Theorem, which states that: \[ (1+x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{k} \] This is the expansion of \( (1+x)^n \) into a sum of terms involving binomial coefficients. ### Deducing that the Sum of Binomial Coefficients Equals \( 2^n \) To deduce that: \[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^{n} \] we can substitute \( x = 1 \) into the binomial expansion: \[ (1+1)^{n} = 2^{n} \] Thus, we have: \[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^{n} \] This completes the proof and the deductions required.