Answer:
[tex]\sum {{n} \atop {1}} \right 6n+2n^2[/tex]
Step-by-step explanation:
Given the series 8 + 12 + 16 + 20+ ...
Sum of nth term Sn = n/2[2a+(n-1)d]
a is the first term = 8
d is the common difference = 12 - 8 = 16 - 12 = 4
Substitute
Sn = n/2[2(8)+(n-1)(4)]
Sn = n/2[16+4n-4]
Sn = n/2[12+4n]
Sn 2n/2[6+2n]
Sn = n(6+2n)
Sn = 6n + 2n²
Hence the sigma representation is expressed as;
[tex]\sum {{n} \atop {1}} \right 6n+2n^2[/tex]
