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 $\left(1+\frac{1}{k}\right)$.

$$\left(1+\frac{1}{k}\right) = \left(\frac{k}{k} + \frac{1}{k}\right) = \frac{k+1}{k}$$

Step 2: Write out the product and identify the pattern.
The product is $P_n = \left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\dots\left(1+\frac{1}{n}\right)$.
Substituting the simplified factors:

$$P_n = \left(\frac{2+1}{2}\right)\left(\frac{3+1}{3}\right)\left(\frac{4+1}{4}\right)\dots\left(\frac{n+1}{n}\right)$$ $$P_n = \left(\frac{3}{2}\right)\left(\frac{4}{3}\right)\left(\frac{5}{4}\right)\dots\left(\frac{n+1}{n}\right)$$

Step 3: Observe the telescoping nature of the product.
The numerator of each term cancels with the denominator of the subsequent term.

$$P_n = \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \dots \times \frac{n+1}{n} = \frac{n+1}{2}$$

Step 4: Determine the values of $n$ for which $P_n$ is an integer.
For $P_n = \frac{n+1}{2}$ 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 $n \ge 2$, the possible values for $n$ are odd integers starting from 3.

$$\boxed{n \in \{3, 5, 7, \dots \}}$$

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 $S_N = \frac{N}{2}[2a + (N-1)d]$.
Substitute the given values:

$$S_{20} = \frac{20}{2}[2(7) + (20-1)5]$$ $$S_{20} = 10[14 + (19)5]$$ $$S_{20} = 10[14 + 95]$$ $$S_{20} = 10[109]$$ $$S_{20} = 1090$$

The sum of this series is $\boxed{1090}$.

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.

$$\boxed{\sum_{k=21}^{75} (5k+2) = 13310}$$

2.2.1

Step 1: Determine the general term for the first differences.
The sequence of first differences is $1; 3; 5; \dots$. 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 $d_n = a_d + (n-1)d_d$.

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