OpenStudy (anonymous):

am i right? A store displays six computers on a shelf side–by–side. If the first computer is eight inches wide and each successive computer is four inches wider than the previous one, find the total width of the computers on the shelf. Part 1: Describe the sigma notation used in answering the question above. (2 points) Part 2: Show all your work and answer the question. (3 points)

6 years ago
OpenStudy (anonymous):

8 ∑ 4 + 4n n=1 ∑ 4 + 4n = 4 ∑ 1 + n 4 * 8 + 4 ∑ n 32 + 4 [8 * 9 / 2] 32 + 40 72 inches

6 years ago
jimthompson5910 (jim_thompson5910):

no, the answer is not \[\Large \sum_{n=1}^{8}4+4n\] for part 1 Part 2 depends on part 1, so that is also incorrect

6 years ago
OpenStudy (anonymous):

ok can you help me then @jim_thompson5910

6 years ago
jimthompson5910 (jim_thompson5910):

How many computers are we talking about here?

6 years ago
OpenStudy (anonymous):

6

6 years ago
jimthompson5910 (jim_thompson5910):

so you're not summing to 8 up at the top (of the sigma), you're summing to 6

6 years ago
OpenStudy (anonymous):

thats the problem above says so i guess so

6 years ago
jimthompson5910 (jim_thompson5910):

so it should be \[\Large \sum_{n = 1}^{6}4+4n\] for part 1

6 years ago
OpenStudy (anonymous):

so is the first and 2nd part for part 2 right?

6 years ago
jimthompson5910 (jim_thompson5910):

first part is, but the second is not...and you'd have to use 6 instead of 8

6 years ago
OpenStudy (anonymous):

∑ 4 + 4n = 4 ∑ 1 + n 4 * 6 + 4 ∑ n 24 + 4 [6 * 7 / 2] 24 + 25 49 inches

6 years ago
OpenStudy (anonymous):

?

6 years ago
jimthompson5910 (jim_thompson5910):

4 [6 * 7 / 2] doesn't become 25

6 years ago
OpenStudy (anonymous):

84? i added thats why

6 years ago
OpenStudy (anonymous):

108 inches ?

6 years ago
jimthompson5910 (jim_thompson5910):

it is 108 inches

6 years ago
jimthompson5910 (jim_thompson5910):

108 is the total final answer for part 2

6 years ago
OpenStudy (anonymous):

can you help me with another problem?

6 years ago
jimthompson5910 (jim_thompson5910):

sure, what is it

6 years ago
OpenStudy (anonymous):

help me with the steps i have an answer just don't know how? geometric partial sum ∑[i=1,5,3(-4)^(i-1)].

6 years ago
OpenStudy (anonymous):

nevermind i found it

6 years ago
jimthompson5910 (jim_thompson5910):

You're basically summing 3(-4)^(i-1) from i = 1 to i = 5 So add up 3(-4)^(i-1) five times like so 3(-4)^(i-1) 3(-4)^(i-1) 3(-4)^(i-1) 3(-4)^(i-1) 3(-4)^(i-1) Then replace the first i with 1, the second i with 2, the third with 3, etc etc like this 3(-4)^(1-1) 3(-4)^(2-1) 3(-4)^(3-1) 3(-4)^(4-1) 3(-4)^(5-1) Now evaluate

6 years ago
OpenStudy (anonymous):

but i do need help with What is the 5th partial sum of ∑[i=1,∞,-1+5n]

6 years ago
OpenStudy (anonymous):

the answer i got for the last problem was 11,170

6 years ago
jimthompson5910 (jim_thompson5910):

same thing, but you're now doing it with -1+5n

6 years ago
OpenStudy (anonymous):

i got 60 ?

6 years ago
jimthompson5910 (jim_thompson5910):

write out -1+5n five times -1+5n -1+5n -1+5n -1+5n -1+5n then replace the first n with 1, the second with 2, etc etc to get -1+5(1) -1+5(2) -1+5(3) -1+5(4) -1+5(5)

6 years ago
jimthompson5910 (jim_thompson5910):

Evaluate each piece, then add them all up

6 years ago
OpenStudy (anonymous):

70?

6 years ago
jimthompson5910 (jim_thompson5910):

yes, you are correct the answer is 70

6 years ago