Evaluation of the Sum from k=1

Evaluation of the Sum ( \Sigma (5k + 8) )

To evaluate the sum represented by the notation ( \Sigma (5k + 8) ) from ( k=1 ) to a specified upper limit, we need to define the range for ( k ) clearly. However, this information appears to be incomplete in the transcript. In order to demonstrate how the evaluation works, let’s assume the upper limit is ( n ). Therefore, the complete sum can be expressed formally as follows:

\sum_{k=1}^{n} (5k + 8)

Step 1: Break the Sum into Two Parts

The sum can be separated into two distinct components:

1. Sum of ( 5k )
2. Sum of ( 8 )

Thus, the expression can be rewritten as:

\sum_{k=1}^{n} (5k + 8) = \sum_{k=1}^{n} 5k + \sum_{k=1}^{n} 8

Step 2: Evaluate Each Component

1. Calculating ( \sum_{k=1}^{n} 5k )
      The term ( 5k ) can be factored out:
   
   
      \sum_{k=1}^{n} 5k = 5 \sum_{k=1}^{n} k
   
      The known formula for the sum of the first ( n ) natural numbers is:
   
   
      \sum_{k=1}^{n} k = \frac{n(n+1)}{2}
   
      Therefore:
   
   
      \sum_{k=1}^{n} 5k = 5 \cdot \frac{n(n+1)}{2} = \frac{5n(n+1)}{2}
   

2. Calculating ( \sum_{k=1}^{n} 8 )
      Since ( 8 ) is a constant, it can be simplified as:
   
   
      \sum_{k=1}^{n} 8 = 8n
   

Step 3: Combine the Results

Now, combining both parts gives:

\sum_{k=1}^{n} (5k + 8) = \frac{5n(n+1)}{2} + 8n

Final Simplification

To further simplify, let’s combine the terms:

1. Finding a common denominator:
   
   
      \frac{5n(n+1)}{2} + \frac{16n}{2} = \frac{5n(n+1) + 16n}{2}
   
2. Thus, we have:
   
   
      \sum_{k=1}^{n} (5k + 8) = \frac{5n^2 + 21n}{2}
   

Conclusion

Therefore, the evaluated sum ( \sum_{k=1}^{n} (5k + 8) ) results in:

\sum_{k=1}^{n} (5k + 8) = \frac{5n^2 + 21n}{2}

This represents the complete evaluation for an unspecified upper limit ( n ). The final result is contingent upon the actual value of ( n ) when it is specified in practical applications.