The nth term of the sequence `1/n, (n+1)/n, (2n+1)/n……..` is

Options

A. `(n^2+1)/n`
B. `(n^2-n+1)/n`
C. n+1
D. 2

Answer

To find the nth term of the sequence given by \( \frac{1}{n}, \frac{n+1}{n}, \frac{2n+1}{n}, \ldots \), we can analyze the pattern in the sequence.

### Step-by-Step Solution:

1. **Identify the Pattern**:
The sequence starts with:
- For \( n = 1 \): \( \frac{1}{1} = 1 \)
- For \( n = 2 \): \( \frac{2+1}{2} = \frac{3}{2} \)
- For \( n = 3 \): \( \frac{3+1}{3} = \frac{4}{3} \)
- For \( n = 4 \): \( \frac{4+1}{4} = \frac{5}{4} \)

From this, we can see that the general form can be expressed as \( \frac{kn + 1}{n} \) for some integer \( k \).

2. **General Term**:
Let's denote the term at position \( k \) as \( T_k \). We can express it as:
\[ T_k = \frac{(k-1)n + 1}{n} \]
Here, \( k \) represents the term number in the sequence.

3. **Finding the nth Term**:
To find the nth term, we substitute \( k = n \):
\[ T_n = \frac{(n-1)n + 1}{n} \]
Simplifying this gives:
\[ T_n = \frac{n^2 - n + 1}{n} = n - 1 + \frac{1}{n} \]

4. **Final Expression**:
Therefore, the nth term of the sequence is:
\[ T_n = n - 1 + \frac{1}{n} \]