(2/15)(3/15)(4/15)(5/15)(6/15)(7/15)(8/15)(9/15... Apr 2026

The product of the sequence is approximately 1. Identify sequence pattern

15!1514the fraction with numerator 15 exclamation mark and denominator 15 to the 14th power end-fraction , which equals approximately (2/15)(3/15)(4/15)(5/15)(6/15)(7/15)(8/15)(9/15...

The expression is a product of fractions.Each numerator increases by 1.The denominator remains constant at 15.The general term is n15n over 15 end-fraction 2. Define the bounds The sequence starts at .Following the pattern, it likely ends at continued to 16, the product would decrease further.If reached 0, the product would be 0. 3. Formulate the product The calculation can be written as: The product of the sequence is approximately 1

∏n=215n15=2⋅3⋅4…151514product from n equals 2 to 15 of n over 15 end-fraction equals the fraction with numerator 2 center dot 3 center dot 4 … 15 and denominator 15 to the 14th power end-fraction 4. Simplify with factorials The numerator is (15 factorial) excluding , the numerator is simply .The denominator is 151415 to the 14th power because there are 14 terms (from ).The formula becomes: # It looks like the product of (n/15) for n starting from 2

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

15!1514the fraction with numerator 15 exclamation mark and denominator 15 to the 14th power end-fraction 5. Calculate numerical value Using the values: Dividing these yields: ≈0.00004479422is approximately equal to 0.00004479422 ✅ Final Answer The product of the sequence from