Question

Can someone please help me with the attached question? It's about finding coefficients of a summation.

Answer 1

looks like a raft of arithmetic to me

Answer 2

I thought it was, and it seemed easy, but I'm obviously doing something wrong because I only got to the second coefficient before I was wrong according to the provided coefficients :(

Answer 3

\[C_7=-\sum_{k=0}^6\frac{C_k}{8-k}\]

Answer 4

way too much arithmetic to write here, you just got to add all that stuff up
why i don't know

Answer 5

So here's my thought process- tell me if something immediately wrong jumps out of you. I know that \[C _{1}\] is 1/2, so \[C _{2}\] would be 1/2+(1/2 over 2+1-1), so 1/2 plus 1/4

Answer 6

\[-\left(\frac{-1}{8}+\frac{1}{12\times 7}+\frac{1}{24\times 6}+\frac{19}{72\times 5}+\frac{3}{360\times 4}+\frac{863}{40480\times 3}\right)\]

Answer 7

they already told you what \(C_2\) is

Answer 8

in your calculator for \(C_2\) you missed the first term with is \(-1\)

Answer 9

Yeah, but that was just a way to check our work. I want to know how to get there. Wait, I'm confused, I'm sorry. So, the firm term is -1, so 1/2 plus 1/4 plus -1?

Answer 10

lets do \(C_2\) and check that it is \frac{1}{12}\) like they say

Answer 11

\[C_2=-\sum_{k=0}^2 \frac{C_k}{3-k}\] right?

Answer 12

I thought it would be \[\sum_{k=0}^{1}\] since the top is n-1

Answer 13

oops
you are right, i am wrong

Answer 14

so this will be even easier

Answer 15

Okay, so I understand so far.

Answer 16

\[C_2=-\sum_{k=0}^2 \frac{C_k}{3-k}\]

Answer 17

\[=-\left(\frac{-1}{3}+\frac{1}{2\times 2}\right)\]

Answer 18

OHMYGOD I get where I went wrong now. THANK YOU!

Answer 19

yw