Question

For the following sequences, find an explicit formula in terms of n for s_n:

1, -(1/2), 1/4, -(1/8), 1/16...

1, 7/9, 10/14, 13/19, 16/24...

1, 3/8, 9/27, 27/64, 81/125...

Answer 1

With s_n being "s sub n"

Answer 2

find the common ratio
for the first one that is -1/2
so
\(\sum_{n=0}^\infty(-\frac{1}{2})^n\)

Answer 3

Yeah, but that doesn't work for s_1, because, in that case, it would be -1/2 instead of 1, right?

Answer 4

is s_n a series or a sequence?

Answer 5

err is it the partial sums of the series ?

Answer 6

No, it's a sequence. Is that the difference? That in a sequence, it doesn't have to work for the first number in the sequence?

Answer 7

By the way, these make absolutely no sense to me, so if I'm missing something remedial, call me out on it.

Answer 8

\(s_n = (-\frac{1}{2})^{n-1}=\{1,-\frac{1}{2},\frac{1}{4},-\frac{1}{8}......\}\)

Answer 9

normally we reserve \(s_n=\sum_{i=1}^n a_n\) where \(a_n \) is a sequence.

Answer 10

Ah. My calculus teacher has s_n on our worksheet.

Answer 11

and you have not started to study series yet?

Answer 12

I don't believe so. There's nothing of series on our worksheet. Then again, our teacher is all over the place at any given time during lecture, so who knows.

Answer 13

I'm just trying to make sure I'm telling you the right thing...

Answer 14

you would know

Answer 15

I believe that series are tomorrow's lecture.

Answer 16

Yes, Series are on Wednesday. Just checked the syllabus.

Answer 17

first we start with sequences \(a_n=a_1,a_2, a_3.....\)
then we start adding them up, and we say
\(s_n=a_1+a_2+a_3+....+a_n\)

Answer 18

ok then forget the first thing I said and go with the second:)

Answer 19

Alright, well, I knew it had to do something with -1/2 from the start, but these other ones I'm completely and utterly lost on.

Answer 20

That's my issue: I can't seem to think of these kind of equations out of the blue.

Answer 21

hmm the 1 is throwing me, but im not the best at these...i just try and construct it.

Answer 22

if the 1 was not there then we could use \(s_n = (\frac{7+3(n-1)}{9+5(n-1)})\)

Answer 23

I can see a pattern, like, "Ok, the numerator is increasing by 3 each time, and the denominator is increasing by 5 each time" but after that I have nothing.

Answer 24

thats exactly how I came up with that^^^

Answer 25

hmm I got to crash because its 1am and I got a final in 8 hours:( but gl.

Answer 26

Alright, good luck and thanks!

Answer 27

Try to describe the pattern using words first, then I'd teach you how to convert words into mathematical notation

Answer 28

Alright, here's my go at the second sequence:

The numerator is increasing by 3 for each iteration, and the denominator is increasing by 5 for each iteration.

Answer 29

Yep, try to do the numerator and the denominator separately.

Answer 30

Alright, give me a sec.

Answer 31

Okay, take your time!

Answer 32

\[(4+3(n-1)) / (4+5(n-1))\]

Answer 33

I'm not very good with the equation editor.

Answer 34

Then simplify it.

Answer 35

You don't need to use the equation editor, just typing it out would do the job

Answer 36

Alright, but as a neat freak, I like it when I can make the equations look nice for others to see. I'll simplify it here in a sec.

Answer 37

(4+3(n−1))/(4+5(n−1)) is as nice as \(\dfrac{4+3(n-1)}{4+5(n-1)}\) :)

Answer 38

I ended up with \[\frac{ 1+3n }{ 5n-1 }\]

Answer 39

Perfect!

Answer 40

YES. I GOT IT TO WORK. (The editor)

Answer 41

P.S. I don't use the editor

Answer 42

Now, about this third one, I got the numerator, but the denominator is throwing me off.

Answer 43

Hint: cubes

Answer 44

Alright. Gimmi a minute to look at it again.

Answer 45

Need the answer?

Answer 46

Nonono, like, one more minute!

Answer 47

Okay, take your time!

Answer 48

\[\frac{ 3^{n-1} }{ n ^{3} }\]

Answer 49

?

Answer 50

Perfect!

Answer 51

Take note of the result generated by the equation editor every time you use it

Answer 52

Soon you'll be able to type equations without the editor!

Answer 53

Yeah, I'm sure if I use it more, it'll be nicer to not have to use it, but for now, I'll take my crutch and run with it.

Answer 54

Err, hobble, maybe.

Answer 55

Thanks for your help, and teaching me what took my teacher a few days, in roughly a half hour or so!

Answer 56

Wow, I'm flattered. No problem!