The sum of an Arithmetic Series is given as:
[tex] S_{n}= \frac{n}{2}(2a_{1}(n-1)*d) [/tex]
where,
d = Common Difference = Difference of any two consecutive terms.
So,
d = 8
a1 = First term = 1
Using the values, we get:
[tex] S_{n}= \frac{n}{2}(2*1+(n-1)*8) \\ \\
S_{n}= \frac{n}{2} (2+8n-8) \\ \\
S_{n}= \frac{n}{2}(8n-6) \\ \\
S_{n}=n(4n-3) \\ \\
S_{n}==4n^{2}-3n [/tex]
The above equation gives the formula for n terms of an Arithmetic Series.
So, option A is the correct answer
