Answer:
1) Σ [n=1, 418] (3+{n-1)4)
Step-by-step explanation:
The expression for the sum is the sum over the number of terms of the expression for a general term.
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The general term of an arithmetic series is ...
an = a1 +d(n -1)
where a1 is the first term, and d is the common difference.
The given series has first term a1=3, and common difference d=7-3 = 4. Then the general term is ...
an = 3 +4(n -1)
The last term will correspond to an n-value that can be found from ...
1671 = 3 +4(n -1)
1668 = 4(n -1) . . . . subtract 3
417 = n -1 . . . . . . . divide by 4
418 = n . . . . . . . add 1
The expression for the sum is ...
[tex]\displaystyle \boxed{\sum_{n=1}^{418}(3+(n-1)4)}[/tex]
