For what the value of k, the following pair of linear equations have infinitely many solutions :
`2x + 3y= 7 and (k+1) x + (2k-1)y= 4k+1`

Answer

To find the value of \( k \) for which the pair of linear equations has infinitely many solutions, we need to use the condition for infinitely many solutions. The two equations given are:

1. \( 2x + 3y = 7 \) (Equation 1)
2. \( (k+1)x + (2k-1)y = 4k + 1 \) (Equation 2)

### Step 1: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1:
- \( a_1 = 2 \)
- \( b_1 = 3 \)
- \( c_1 = 7 \)

- For Equation 2:
- \( a_2 = k + 1 \)
- \( b_2 = 2k - 1 \)
- \( c_2 = 4k + 1 \)

### Step 2: Set up the condition for infinitely many solutions
For the two equations to have infinitely many solutions, the following condition must hold:

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

### Step 3: Set up the equations
Using the identified coefficients, we can set up the equations:

1. From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \):
\[ \frac{2}{k+1} = \frac{3}{2k-1} \]

2. From \( \frac{a_1}{a_2} = \frac{c_1}{c_2} \):
\[ \frac{2}{k+1} = \frac{7}{4k+1} \]

### Step 4: Solve the first equation
Cross-multiplying the first equation:
\[ 2(2k - 1) = 3(k + 1) \]
Expanding both sides:
\[ 4k - 2 = 3k + 3 \]
Rearranging gives:
\[ 4k - 3k = 3 + 2 \]
\[ k = 5 \]

### Step 5: Verify with the second equation
Now, let's verify if \( k = 5 \) satisfies the second condition:
Substituting \( k = 5 \) into the second equation:
\[ \frac{2}{5 + 1} = \frac{7}{4(5) + 1} \]
Calculating both sides:
Left side:
\[ \frac{2}{6} = \frac{1}{3} \]
Right side:
\[ \frac{7}{20 + 1} = \frac{7}{21} = \frac{1}{3} \]
Both sides are equal, confirming that \( k = 5 \) satisfies both conditions.

### Conclusion
Thus, the value of \( k \) for which the pair of linear equations has infinitely many solutions is:

\[ \boxed{5} \]