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anonymous
 one year ago
Does anyone know how to do this Sigma Notation problem?
anonymous
 one year ago
Does anyone know how to do this Sigma Notation problem?

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1it starts at i=5 and ends at i=13 you do replace i with 5 + sign replace i with 6 + sign replace i with 7 + sign replace i with 8 + sign replace i with 9 + sign replace i with 10 + sign replace i with 11 + sign replace i with 12 + sign replace i with 13 end

freckles
 one year ago
Best ResponseYou've already chosen the best response.1then you just do order of operations

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\text{ Let } S_{i=1 \text{ to } i=13}=\sum_{i=1}^{13}(3i+2)= \sum_{i=1}^{13}(3i) + \sum_{i=1} ^{13} 2 \\ S_{ i=1 \text{ to } i=13 }=3 \frac{13(13+1)}{2}+2(13) \] Bur we don't want to include what happens at i=1 or i=2 or i=3 or i=4 so we want to exclude the sum from i=1 to i=4 \[S_{i=5 \text{ to } i=13}=S_{i=1 \text{ to } i=13}S_{i=1 \text{ to } i=4} \] where \[S_{i=1 \text{ to } i=4} =\sum_{i=1}^4 (3i+2)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1can you tell me on what part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The last part that you added

freckles
 one year ago
Best ResponseYou've already chosen the best response.1like are you talking about applying the following formulas: \[\sum_{i=1}^{n} i=\frac{n(n+1)}{2} \\ \text{ and } \\ \sum_{i=1}^n c=cn\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1or are you talking about \[\sum_{i=1}^{13} f(i)=\sum_{i=1}^4 f(i)+\sum_{i=5}^{13} f(i) \\\ \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, the part where you said Si=1 to I=4

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I don't quite understand if you understand the last thing I asked you about then why are you asking me about the sum from i=1 to i=4

freckles
 one year ago
Best ResponseYou've already chosen the best response.1c=a+b is equivalent to the equation ca=b or also known as b=ca

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to understand what you're saying ...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum_{i=1}^{13}f(i)=\sum_{i=1}^{4} f(i)+\sum_{i=5}^{13} f(i) \\ \text{ subtract } \sum_{i=1}^4 f(i) \text{ on both sides } \\ \sum_{i=1}^{13} f(i)\sum_{i=1}^4f(i)=\sum_{i=5}^{13} f(i)\] as you see the sum from i=5 to i=13 is equal to difference of (the sum from i=1 to i=13) and (the sum from i=1 to i=4)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand anything you're typing, but thanks anyway.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1hmmm... I wonder how they expect you to evaluate the thing involving sigma if you never seen it before

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum_{i=1}^{13} f(i)=f(1)+f(2)+f(3)+f(4)+ f(5)+f(6)+f(7)+f(8)+f(9)+ \\ f(10)+f(11)+f(12)+ f(13) \\ \\ \sum_{i=1}^{4} f(i)=f(1)+f(2)+f(3)+f(4) \\ \text{ as you see subtract the second line from the first }\\ \text{ (first actually occupies two lines because I couldn't fit it on one) gives } \\ f(5)+f(6)+f(7)+f(8)+f(9)+f(10)+f(11)+f(12)+f(13) \\ \text{ which is the same as } \sum_{i=5}^{13} f(i)\]
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