Question

Karama's gauntlet of calc questions 1) using the first four terms of a series, find the series sum:

Answer 1

@wio

Answer 2

this isn't the same one I believe.

Answer 3

did you try 9?

Answer 4

it is just a hunch btw do not crucify me if it is wrong

Answer 5

What is the current topic of the course?

Answer 6

series, in order to start integral approaches to series, as well as sequences

Answer 7

You have to do a bit of algebra first of all.

Answer 8

\[
\sum [f(n)+g(n)] = \sum f(n)+\sum g(n)
\]

Answer 9

We'll start with:\[

\large
\sum_{n=0}^\infty \frac{3}{2^n}
\]

Answer 10

@karama following so far?

Answer 11

so that is my first term in the compound functioN?

Answer 12

First we pull out the \(0\)th term.\[
\large
\sum_{n=0}^\infty \frac{3}{2^n} = 3+\sum_{n=1}^\infty \frac{3}{2^{n}}
\]

Answer 13

pull out? I like that viewpoint, very cool!

Answer 14

Next we factor out a \(2\) so we have \(n-1\)\[
\large
3+\sum_{n=1}^\infty \frac{3}{2^{n}} = 3+\sum_{n=1}^\infty \frac{3}{2}\frac{1}{2^{n-1}}
\]

Answer 15

and is the 3/2 just rearranging the product on top?

Answer 16

Now if we let \(a = 3/2\) and \(r=1/2\):\[

\large
3+\sum_{n=1}^\infty \frac{3}{2}\frac{1}{2^{n-1}} = 3+\sum_{n=1}^\infty ar^{n-1} = \frac{a}{1-r} = \frac{3/2}{1-(1/2)}
\]

Answer 17

wait, there should be a \(3+\) in front of all of them.

Answer 18

@karama think you can do it now?

Answer 19

ah!!! my professor was talking about a problem like this the other day. When he pulled out a C we all did a double take. Yeah, I think I have it. Thanks so much!

Answer 20

And thanks sumner for helping me out too! I really appreciate it :)