Simplify the product (1+(1)/(2))(1+(1)/(3))(1+(1)/(4))(1+(1)/(n)) and determine the values of n for which the result is an integer.

Here are the solutions to the problems: 1.3 Step 1: Simplify each factor in the product. Each factor is of the form (1+(1)/(k)). (1+(1)/(k)) = ((k)/(k) + (1)/(k)) = (k+1)/(k) Step 2: Write out the product and identify the pattern. The product is P_n = (1+(1)/(2))(1+(1)/(3))(1+(1)/(4))(1+(1)/(n)). Substituting the simplified factors: P_n = ((2+1)/(2))((3+1)/(3))((4+1)/(4))((n+1)/(n)) P_n = ((3)/(2))((4)/(3))((5)/(4))((n+1)/(n)) Step 3: Observe the telescoping nature of the product. The numerator of each term cancels with the denominator of the subsequent term. P_n = (3)/(2) × (4)/(3) × (5)/(4) × × (n+1)/(n) = (n+1)/(2) Step 4: Determine the values of n for which P_n is an integer. For P_n = (n+1)/(2) to be an integer, n+1 must be an even number. If n+1 is even, then n must be an odd number. Given that n 2, the possible values for n are odd integers starting from 3. n \3, 5, 7, \ 2.1.1 Step 1: Identify the given values for the arithmetic series. First term, a = 7. Common difference, d = 5. Number of terms, N = 20. Step 2: Use the formula for the sum of an arithmetic series. The sum of an arithmetic series is S_N = (N)/(2)[2a + (N-1)d]. Substitute the given values: S_20 = (20)/(2)[2(7) + (20-1)5] S_20 = 10[14 + (19)5] S_20 = 10[14 + 95] S_20 = 10[109] S_20 = 1090 The sum of this series is 1090. 2.1.2 Step 1: Calculate the sum of the terms added to the original series. The original series has 20 terms, and its sum is S_20 = 1090. The extended series has 75 terms, and its sum is S_75 = 14400. The sum of the terms added is S_added = S_75 - S_20. S_added = 14400 - 1090 = 13310 Step 2: Determine the general term T_k of the arithmetic series. The general term is T_k = a + (k-1)d. T_k = 7 + (k-1)5 T_k = 7 + 5k - 5 T_k = 5k + 2 Step 3: Write the equation for the sum of the added terms using sigma notation. The terms added are from the 21st term to the 75th term. _k=21^75 (5k+2) = 13310 2.2.1 Step 1: Determine the general term for the first differences. The sequence of first differences is 1; 3; 5; . This is an arithmetic progression. First term of differences, a_d = 1. Common difference of differences, d_d = 2. The n-th term of the first differences is d_n = a_d + (n-1)d_d. d_n = 1 + (n