Answer:
[tex]\Sigma(2n-1) \left \{ {{n=50} \atop {n=1}} \right. \hspace{5} n\in N[/tex]
Step-by-step explanation:
First, we need to find the sequence associated to the summation. As you can see the sum only takes odd numbers into account. Hence the sequence it is defined by:
[tex]2n-1[/tex]
Where:
[tex]n\in N[/tex]
Now, we only need to find the limits of the summation:
Evaluating n=1
[tex]2(1)-1=2-1=1[/tex]
Evaluating n=50
[tex]2(50)-1=100-1=99[/tex]
Therefore the summation can be written as:
[tex]\Sigma(2n-1) \left \{ {{n=50} \atop {n=1}} \right.[/tex]
