Question

Does anyone know how to do this Sigma Notation problem?

Answer 1

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

Answer 2

then you just do order of operations

Answer 3

or ...

Answer 4

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

Answer 5

I'm confused

Answer 6

can you tell me on what part

Answer 7

The last part that you added

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

\[\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)\]