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

1. anonymous

2. freckles

it 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

3. freckles

then you just do order of operations

4. freckles

or ...

5. freckles

$\text{ Let } S_{i=1 \text{ to } i=13}=\sum_{i=1}^{13}(3i+2)= \sum_{i=1}^{13}(3i) + \sum_{i=1} ^{13} 2 \\ S_{ i=1 \text{ to } i=13 }=3 \frac{13(13+1)}{2}+2(13)$ 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 $S_{i=1 \text{ to } i=4} =\sum_{i=1}^4 (3i+2)$

6. anonymous

I'm confused

7. freckles

can you tell me on what part

8. anonymous

The last part that you added

9. freckles

like are you talking about applying the following formulas: $\sum_{i=1}^{n} i=\frac{n(n+1)}{2} \\ \text{ and } \\ \sum_{i=1}^n c=cn$

10. freckles

or are you talking about $\sum_{i=1}^{13} f(i)=\sum_{i=1}^4 f(i)+\sum_{i=5}^{13} f(i) \\\$

11. anonymous

No, the part where you said Si=1 to I=4

12. freckles

I 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

13. freckles

c=a+b is equivalent to the equation c-a=b or also known as b=c-a

14. anonymous

I'm trying to understand what you're saying ...

15. freckles

$\sum_{i=1}^{13}f(i)=\sum_{i=1}^{4} f(i)+\sum_{i=5}^{13} f(i) \\ \text{ subtract } \sum_{i=1}^4 f(i) \text{ on both sides } \\ \sum_{i=1}^{13} f(i)-\sum_{i=1}^4f(i)=\sum_{i=5}^{13} f(i)$ 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)

16. anonymous

I don't understand anything you're typing, but thanks anyway.

17. freckles

hmmm... I wonder how they expect you to evaluate the thing involving sigma if you never seen it before

18. freckles

$\sum_{i=1}^{13} f(i)=f(1)+f(2)+f(3)+f(4)+ f(5)+f(6)+f(7)+f(8)+f(9)+ \\ f(10)+f(11)+f(12)+ f(13) \\ \\ \sum_{i=1}^{4} f(i)=f(1)+f(2)+f(3)+f(4) \\ \text{ as you see subtract the second line from the first }\\ \text{ (first actually occupies two lines because I couldn't fit it on one) gives } \\ f(5)+f(6)+f(7)+f(8)+f(9)+f(10)+f(11)+f(12)+f(13) \\ \text{ which is the same as } \sum_{i=5}^{13} f(i)$