Question

Pre-calculus Honors. Sigma :)

Answer 1

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Answer 2

is it a test @43v3rFr3sh ?

Answer 3

Okay, you guys ready? Its pretty simple, I just cant comprehend it. lol

Evaluate the summation of 3 n plus 2, from n equals 1 to 14.

Answer 4

No, its definitely not 6. lol

Answer 5

@AlexandervonHumboldt2 No, its just a homework problem.

Answer 6

Slow Way

Let f(n) = 3n+2

Plug in n = 1 to get f(1) = 3*1+2 = 3+2 = 5
Plug in n = 2 to get f(2) = 3*2+2 = 6+2 = 8
etc etc

you repeat these steps all the way up to n = 14

Once you have your 14 outputs, you add them all up

------------------------------------------------

Faster way

\[\Large \sum_{n=1}^{14}\left(3n+2\right)\]
\[\Large \sum_{n=1}^{14}\left(3n\right)+\sum_{n=1}^{14}\left(2\right)\]
\[\Large 3*\sum_{n=1}^{14}\left(n\right)+\sum_{n=1}^{14}\left(2\right)\]
Now you'll use these formulas
\[\Large \sum_{k=1}^{n}\left(k\right) = \frac{n(n+1)}{2}\]
\[\Large \sum_{k=1}^{n}\left(C\right) = C*n\]
where C is some constant

Answer 7

What if my teacher gives me a problem with 54 terms in the summation?

Answer 8

then I suggest you use the second method (the fast way) instead of generating 54 terms and adding them up

Answer 9

Can you guide me through that way, using an example problem? I don't really understand where to start.

Answer 10

ok let's say we had
\[\Large \sum_{n=1}^{54}\left(7n+5\right)\]

Answer 11

Okay. :)

Answer 12

First we split things up like so
\[\Large \sum_{n=1}^{54}\left(7n+5\right)\]
\[\Large \sum_{n=1}^{54}\left(7n\right)+\sum_{n=1}^{54}\left(5\right)\]
\[\Large 7*\sum_{n=1}^{54}\left(n\right)+\sum_{n=1}^{54}\left(5\right)\]
Hopefully you're able to see how I did these steps so far?

Answer 13

Yes, I understand so far. :)

Answer 14

ok great

Answer 15

Now let's focus on each sum (sigma) one at a time
First up is \[\Large \sum_{n=1}^{54}\left(n\right)\]

Answer 16

Okay

Answer 17

what does this sum actually mean? any ideas?

Answer 18

The adding of a series of continuity? Potentially exponentially.

Answer 19

?

Answer 20

see how we have 'n = 1' at the bottom? and 54 at the top?

those two mean "start the sum at n = 1 and end it at n = 54"

we will add up 54 different values
n = 1
n = 2
n = 3
....
....
....
n = 52
n = 53
n = 54

We could use the slow method to actually add up these 54 terms, or we can use this shortcut

\[\Large \sum_{n=1}^{m}\left(n\right) = \frac{m(m+1)}{2}\]

in this case, m = 54, so

\[\Large \sum_{n=1}^{54}\left(n\right) = \frac{54(54+1)}{2} = 1,485\]