1. The problem is to find the 7th row of Pascal's triangle. 2. Pascal's triangle rows are numbered starting from 0, where the 0th row is [1]. The nth row contains the binomial coefficients $ \binom{n}{k} $ for $ k=0,1,\ldots,n $. 3. The formula for the entries in the nth row is: $$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$ where $ n! $ is the factorial of n. 4. For the 7th row (where $ n=7 $), calculate each element $ \binom{7}{k} $ for $ k=0 $ to $ 7 $: - $ \binom{7}{0} = \frac{7!}{0!7!} = 1 $ - $ \binom{7}{1} = \frac{7!}{1!6!} = 7 $ - $ \binom{7}{2} = \frac{7!}{2!5!} = 21 $ - $ \binom{7}{3} = \frac{7!}{3!4!} = 35 $ - $ \binom{7}{4} = \frac{7!}{4!3!} = 35 $ - $ \binom{7}{5} = \frac{7!}{5!2!} = 21 $ - $ \binom{7}{6} = \frac{7!}{6!1!} = 7 $ - $ \binom{7}{7} = \frac{7!}{7!0!} = 1 $ 5. Therefore, the 7th row of Pascal's triangle is: $$ [1, 7, 21, 35, 35, 21, 7, 1] $$ This row represents the coefficients of the expansion of $ (a+b)^7 $.