Question

given the image of pascals triangle below, if you kept writing out the triangle what would be the values for the n=9 row.

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1

select one:
○ a. 1, 8, 28, 56, 70, 56, 28, 8, 1
○ b. 1, 9, 81, 729, 81, 9, 1
○ c. 1, 9, 36, 84, 126, 126, 84, 36, 9, 1
○ d. 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1

Explanation:

Step1: Recall Pascal's Triangle Rule

Each row \( n \) (starting from \( n = 0 \)) of Pascal's Triangle has entries given by binomial coefficients \( \binom{n}{k} \) for \( k = 0, 1, \dots, n \), and each entry (except the first and last) is the sum of the two entries above it. Also, row \( n \) has \( n + 1 \) elements. For \( n = 9 \), we need to find the 10th row (since \( n = 0 \) is the first row).

Step2: Calculate Binomial Coefficients for \( n = 9 \)

The binomial coefficient is given by \( \binom{n}{k}=\frac{n!}{k!(n - k)!} \).

• For \( k = 0 \): \( \binom{9}{0}=\frac{9!}{0!9!}=1 \)
• For \( k = 1 \): \( \binom{9}{1}=\frac{9!}{1!8!}=9 \)
• For \( k = 2 \): \( \binom{9}{2}=\frac{9!}{2!7!}=\frac{9\times8}{2\times1}=36 \)
• For \( k = 3 \): \( \binom{9}{3}=\frac{9!}{3!6!}=\frac{9\times8\times7}{3\times2\times1}=84 \)
• For \( k = 4 \): \( \binom{9}{4}=\frac{9!}{4!5!}=\frac{9\times8\times7\times6}{4\times3\times2\times1}=126 \)
• For \( k = 5 \): By symmetry of binomial coefficients (\( \binom{n}{k}=\binom{n}{n - k} \)), \( \binom{9}{5}=\binom{9}{4}=126 \)
• For \( k = 6 \): \( \binom{9}{6}=\binom{9}{3}=84 \)
• For \( k = 7 \): \( \binom{9}{7}=\binom{9}{2}=36 \)
• For \( k = 8 \): \( \binom{9}{8}=\binom{9}{1}=9 \)
• For \( k = 9 \): \( \binom{9}{9}=\binom{9}{0}=1 \)

So the \( n = 9 \) row (10th row) is \( 1, 9, 36, 84, 126, 126, 84, 36, 9, 1 \), which matches option c.

Answer:

c. 1, 9, 36, 84, 126, 126, 84, 36, 9, 1