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Pascal's Triangle is a triangular array of numbers in which each number is the sum of the two numbers directly above it. It starts with a row of 1 on the top and builds the subsequent rows using the following rule: Each entry of a row is obtained by adding the two entries above it in the previous row. The first and last entries of each row are always 1. Here are the first few rows of Pascal's Triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1

Let's calculate the sum of the entries for each of the given rows: Row n+1 = 1: Sum = 1 = 2^0 Row n+1 = 2: Sum = 1 + 1 = 2 = 2^1 Row n+1 = 3: Sum = 1 + 2 + 1 = 4 = 2^2 Row n+1 = 4: Sum = 1 + 3 + 3 + 1 = 8 = 2^3 Row n+1 = 5: Sum = 1 + 4 + 6 + 4 + 1 = 16 = 2^4 Row n+1 = 6: Sum = 1 + 5 +10 +10 + 5 + 1= 32 = 2^5

Now, we'll use mathematical induction to prove that the sum of entries in row n+1 of Pascal's Triangle equals 2^n for any non-negative integer n. Step 3.1: Base Case: We already calculated the sum of the entries for n = 0, 1, 2, 3, 4, and 5. For each of these values, the sum is equal to 2^n. Thus, our base case holds. Step 3.2: Inductive Step: Assume that the sum of entries in row n+1 is equal to 2^n. We need to prove that the sum of entries in row (n+1) + 1 is equal to 2^(n+1). For each entry in row n+2, we can write it as the sum of the two entries above it in row n+1. Let S represent the sum of the entries in row n+2. Then, S = (entry 1 in row (n+1)) + (entry 2 in row (n+2)) + ... + (entry n+2 in row (n+2)) Now, each of these intermediate sums corresponds to adding two numbers from row n+1. For example, entry 2 in row (n+2) = entry 1 in row (n+1) + entry 2 in row (n+1) So, we can rewrite the sum S by adding the entries in row n+1 twice: S = 2 × (sum of entries in row (n+1)) Since we've assumed the sum of entries in row n+1 is equal to 2^n, we get: S = 2 × 2^n = 2^(n+1) Thus, the sum of entries in row (n+1) + 1 is equal to 2^(n+1). This completes the inductive step. By mathematical induction, we have now proven that the sum of the entries in row n+1 of Pascal's Triangle equals 2^n for any non-negative integer n.