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find the sum of each finite series
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\[\sum_{n=1}^{10}(3n-4)\]
what do you get when n = 1?
what do you get for n = 2? . . what do you get for n = 10? then add those numbers up. There is your sum
12?
where did that 12 come from?
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\(\sum_{n=1}^{10}(3n-4)=3\sum_{n=1}^{10}n-\sum_{n=1}^{10}4=3\sum_{n=1}^{10}n-40\)
Might I recommend using the formula \[[n(a _{1} + a _{n})]\div2\]
Where n is the number of terms, a1 is the first term, and an is the last term.
right,or you can use the short cut like others just recommended. (number of terms) (first term + last term)/2
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