The common difference in the arithmetic sequence is d = 6 (each term increases by 6 compared to the previous term). The first term is a_1 = 7.
To write the sum of the first 51 terms using sigma notation with the lower limit n=2, we need to find the upper limit of the summation. We can use the formula for the nth term of an arithmetic sequence:
a_n = a_1 + (n-1)d
Setting a_n equal to the last term we want to include in the sum (which is the 51st term), we get:
a_51 = 7 + (51-1)6 = 307
So the upper limit of the summation is n = 51.
Now we can use sigma notation to write the sum of the first 51 terms:
∑(n=2 to 51) (7 + (n-1)6)
This notation means that we start the summation from n = 2 and add up the terms in the sequence for n = 2, 3, 4, ..., 51. The term inside the parentheses represents each term in the sequence, which is 7 plus the common difference multiplied by (n-1).
To evaluate this summation, we can simplify the expression inside the parentheses:
7 + (n-1)6 = 6n + 1
Now we can substitute this expression back into the sigma notation:
∑(n=2 to 51) (6n + 1)
Expanding the summation, we get:
(6(2) + 1) + (6(3) + 1) + (6(4) + 1) + ... + (6(51) + 1)
Simplifying, we get:
13 + 19 + 25 + ... + 307
To find the sum of this arithmetic series, we can use the formula:
S_n = (n/2)(a_1 + a_n)
where S_n is the sum of the first n terms, a_1 is the first term, and a_n is the nth term.
Plugging in the values, we get:
S_51 = (51/2)(7 + 307) = 7863
Therefore, the sum of the first 51 terms of the arithmetic sequence is 7863.