Question

Write the sum using summation notation, assuming the suggested pattern continues. 2 - 12 + 72 - 432 + ...

Answer 1

Have you found anything so far? Or not sure how to begin?

Answer 2

Not sure how to begin lol

Answer 3

Ah, no problem! First, we should figure out the pattern that is going on. Summation notation can wait a bit!
Can you see what is going on between each next term, from 2 to -12, to 72, and to -432 ?

Answer 4

That's a nice thing to check first.
From 2 to -12, it does seem like we might have subtraction! Subtracting 14 from 2 to get -12.
But does the pattern continue? -12 to 72, that -14 pattern seems to break!
Perhaps instead, there is something being multiplied instead?

Answer 5

from -12 to 72 it's 84

Answer 6

True, but a pattern is saying that we have a common step that continues in every term.
Like, 1, 2, 3, 4, 5 is a pattern of adding 1's. 9, 4, -1, -6, ... is a pattern of subtracting 5's. 1, 3, 9, 27, 81... is a pattern of multiplying by 3's. In each case, you get from one part to the next by using the pattern.

Does that make sense?

Answer 7

So, there are two main patterns worked with very often in math: arithmetic (adding/subtracting a common number) and geometric (multiplying/dividing a common number). These are the two we must always check for summations.

You did the check for arithmetic by subtracting the next number by the previous. -12 - 2 = -14. So our pattern, if any, should be -14. But 72 - -12 = +84, which does not follow the pattern. It would have to be -12 - 14 = -26 to follow that pattern, followed by -40 and so on. That's not what we have!

So we can check for a geometric pattern instead by dividing the next number by the previous to see if there is a common factor. For example:
2, -12, 72, -432, ...
(1) -12 / 2 = -6.
(2) 72 / -12 = -6
(3) -432 / 72 = -6
It appears we have a pattern of multiplying by -6.

Answer 8

Ahh okay

Answer 9

\[\sum_{n=0}^{\infty} 2(-6)^(n-1) \]

Answer 10

(-6)^(n-1)

Answer 11

Looks close. But if we plug in the first index, n=0, that's 2*(-6)^(0-1) = 2*(-6)^(-1). That's not the first term. (This is a nice check to do as well, it's fairly easy in many cases to plug in and check if the first term matches the first index).
So we could do one of two things:
change the index to n=1 instead of n=0,
or change (-6)^(n-1) for (-6)^n.

Answer 12

(-6)^n

Answer 13

Alright, so if we make that change:
\( \displaystyle \quad \sum_{n=0}^{\infty} 2(-6)^n \)
If we plug in n=0 now, 2*(-6)^ 0 = 2*1 = 2. Looks good now!

Answer 14

Yay!
Thank you((:

Answer 15

Yup, glad to help! :)