(2/15)(3/15)(4/15)(5/15)(6/15)(7/15)(8/15)(9/15... Here

15!1514the fraction with numerator 15 exclamation mark and denominator 15 to the 14th power end-fraction , which equals approximately

The product of the sequence is approximately 1. Identify sequence pattern (2/15)(3/15)(4/15)(5/15)(6/15)(7/15)(8/15)(9/15...

import math # Extracting the sequence from the subject # (2/15)(3/15)(4/15)(5/15)(6/15)(7/15)(8/15)(9/15... # It looks like the product of (n/15) for n starting from 2. # Let's assume it goes up to 14 or 15. def calculate_product(limit): product = 1 for i in range(2, limit + 1): product *= (i / 15) return product results = {limit: calculate_product(limit) for limit in range(9, 16)} print(results) Use code with caution. Copied to clipboard (2/15)(3/15)(4/15)(5/15)(6/15)(7/15)(8/15)(9/15...