The sum of first n numbers of the form `(5k + 1)`, where `k in I^+` is :

Options

A. `n/2 [5n^2 - 3]`
B. `n (20 - 3n)`
C. `n/2 (5n + 7)`
D. none of these

Answer

To find the sum of the first \( n \) numbers of the form \( (5k + 1) \) where \( k \) is a positive integer, we can follow these steps:

### Step 1: Identify the first few terms
The first few terms of the sequence can be generated by substituting positive integer values for \( k \):
- For \( k = 1 \): \( 5(1) + 1 = 6 \)
- For \( k = 2 \): \( 5(2) + 1 = 11 \)
- For \( k = 3 \): \( 5(3) + 1 = 16 \)
- For \( k = 4 \): \( 5(4) + 1 = 21 \)

Thus, the first \( n \) terms are:
- \( 6, 11, 16, 21, \ldots \)

### Step 2: Recognize the pattern
The sequence \( 6, 11, 16, 21, \ldots \) is an arithmetic progression (AP) where:
- The first term \( a = 6 \)
- The common difference \( d = 5 \)

### Step 3: Use the formula for the sum of the first \( n \) terms of an AP
The formula for the sum \( S_n \) of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]

### Step 4: Substitute the known values into the formula
Substituting \( a = 6 \) and \( d = 5 \) into the formula:
\[ S_n = \frac{n}{2} \times (2 \times 6 + (n - 1) \times 5) \]

### Step 5: Simplify the expression
Calculating the terms inside the parentheses:
\[ 2 \times 6 = 12 \]
\[ (n - 1) \times 5 = 5n - 5 \]
Thus, we have:
\[ S_n = \frac{n}{2} \times (12 + 5n - 5) \]
\[ S_n = \frac{n}{2} \times (5n + 7) \]

### Step 6: Final expression for the sum
Therefore, the sum of the first \( n \) numbers of the form \( (5k + 1) \) is:
\[ S_n = \frac{n(5n + 7)}{2} \]

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