Final answer:
The first 9 terms of the sequence are 2, 3, 13, 18, 23, 28, 33, 38, and 43. The recursive relation in the question seems incorrect, but if we follow the assumption (S_n = 5n - 2), we find that the sum of the first n terms of the sequence is ¾ n², which is derived from the arithmetic series sum formula.
Step-by-step explanation:
First, let's list the first 9 terms of the recursively defined sequence, which is initialized with (S_1 = 2) and (S_2 = 3). The recurrence relation given is (S_n = 5n - 2) and (S_{n-1} - 1). However, this recurrence seems to be incorrectly stated because it defines two separate forms of (S_n) and doesn't give a proper way to derive (S_n) from previous terms. Instead, we'll derive the first 9 terms using the initial values and the assumption that (S_n = 5n - 2) since no proper recursive formula is provided:
- S_1 = 2
- S_2 = 3
- S_3 = 5(3) - 2 = 13
- S_4 = 5(4) - 2 = 18
- S_5 = 5(5) - 2 = 23
- S_6 = 5(6) - 2 = 28
- S_7 = 5(7) - 2 = 33
- S_8 = 5(8) - 2 = 38
- S_9 = 5(9) - 2 = 43
To solve the second part, the sum of the first n terms of this sequence, which we now know is given by (S_k = 5k - 2), can be found by summing each term individually or by finding a formula for the sum. The sum is a sum of an arithmetic series:
(sum_{k=1}^n S_k) = S_1 + S_2 + ... + S_n
For the given sequence, if we assume n terms, and that the expression in the box is equal to n², we have a situation similar to a proof technique where we pair terms together. Though the example provided involves operations like taking (n - 1) from the last term and so on, which might not directly apply to our arithmetic sequence. If we proceeded with an arithmetic series, we could use the formula for the sum of an arithmetic series:
S_n = ⅓ n (first term + last term) = ⅓ n (S_1 + S_n)
Substitute the expressions of S_1 and S_n:
= ⅓ n (2 + (5n - 2))
= ⅓ n (5n) = ½ n(5n) = ¾ n²
The series sum formula gives us a clear path to calculate the sum directly, rather than using the illustration provided with the taking and adding terms.