Question

Find the general term for the sequence:
-1/6, 1/36, -1/108, 1/216...

Answer 1

I get that it would include (-1)^n+1 to account for the alternating sign. I am stuck past that

Answer 2

the problem I am really working on is trying to find the taylor series for \[\frac{ 1 }{ 1-x } \] centered at c=7, if that helps for any reason

Answer 3

I know the maclaurin series for 1/1-x, but not when the center gets shifted

Answer 4

So let's go with what you know. What's the MacLaurin Series for \[(\frac{ 1 }{ 1-x })\]?

Answer 5

And we'll go from there!

Answer 6

\[\sum_{n=0}^{\infty} x ^{n}\]

Answer 7

Okay, awesome! So do you know what "phase shift" in a function is? Like trig. functions?

Answer 8

yes, I think so
I tried just putting (x-7) inside the summation, but that wasn't correct apparently

Answer 9

meaning \[\sum_{n=0}^{\infty} (x-7)^{n}\]

Answer 10

Okay, well, technically, your MacLaurin series was correct, but here is the general form for all MacLaurin Series: \[\sum_{n=0}^{\infty} \] \[\frac{ {f}^{n}(0) }{ n! }(x-0)^n\]

Answer 11

A Taylor Series, is just taken by replacing those 0's with values [aka, were your series is centered at]

Answer 12

I get that, my problem then is finding the nth derivative of 7. thats what i was trying to find out with the question i asked at the start

Answer 13

SO, the Taylor Series General form is \[\sum_{n}^{\infty} \frac{ f^n(a) }{n! }(x-a)^n\]

Answer 14

Okay, so in this case, the a=7, correct?

Answer 15

yes

Answer 16

I have to leave, but I will defintely look back at this if you feel like still working on it. I really appreciate it!

Answer 17

Okay, sorry I am taking so long! Who said that the (x-7)^n was wrong?

Answer 18

OKAY, so here's the trick! What you need to do, is you need to actually start listing out those derivatives [as painful as it's gonna be]! Here we go!

f(7)=(1/[1-7])=-(1/6)
f'(7)=(1/{[1-7]^2})=1/36
f''(7)=(1/[1-7])^3=-(1/216)
etc.
\[f^{(n)} = (\frac{ -1 }{6 })^n\]

THEREFORE: The Taylor Series simplifies down to:

\[\sum_{n=0}^{\infty} (\frac{ -1 }{ 6 })^n(x-7)^n\]
Hope this helped!!!