Sequences and series - try again

Click For Summary

Homework Help Overview

The discussion revolves around sequences and series, specifically focusing on the k-th term and the k-th partial sum of two geometric series. The original poster expresses confusion regarding the problem's requirements and the appropriate approach to find the largest natural number k where the difference between the terms of the two series exceeds 14.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting up an inequality to solve for k and explore the implications of the k-th term versus the k-th partial sum. There is uncertainty about whether to subtract the second series' partial sum from the first or to focus on the terms themselves.

Discussion Status

Some participants have provided guidance on setting up the problem, while others express frustration with the question's clarity. There is an ongoing exploration of the definitions and terms involved, particularly regarding the distinction between terms and partial sums in geometric series.

Contextual Notes

Participants note confusion stemming from the terminology used in the problem statement, particularly the phrase "k-th term of the series" versus "k-th partial sum." This has led to varied interpretations of what is being asked.

oilersforever72
Messages
7
Reaction score
0
Sequences and series - try again :)

Hi, I'm going to try to post this question again, hopefully it is more clear this time. I'm not sure how to approach this question, or really, what this question is asking me!

Homework Statement



The k-th term of a series, Sk = a*[(1-(r^k))/(1-r)], is the sum of the first k terms of the underlying sequence.

(Note: This is a general formula that I remember from grade 12 math where a is the first term in a sequence, and r is the constant ratio between subsequent terms. Correct me if I'm wrong :smile:.)

The difference between the n-th terms of two particular series is greater than 14 for some values of n (where n is a Natural number). The series with general term tn = 100[(11/17)^(n-1)] begins larger than the second series with general term tn = 50[(14/17)^(n-1)]. Find the largest natural number, k, where the difference between the terms of these two series is larger than 14.

Homework Equations





The Attempt at a Solution



I'm hooped.
 
Last edited:
Physics news on Phys.org
Try to set up an inequality you hope can be solved for k! :smile:
 
Hmm... Okay, what is the problem actually asking for?

FIRST Sk: So given the first general term, I can plug it into the Sk formula to get Sk = 100[1-(11/17)^k)]/[(1-(11/17)].

SECOND Sk: Given the second general term, I can plug it into the Sk formula again to get Sk = 50[1-(14/17)^k]/[1-(14/17)].

Is it asking me to solve for k by subtracting the second Sk from the first Sk and setting it equal or greater to 14?
 
Hmm...
 
Indeed it is!
However, I know of any simple formula to compute this, with a subtraction between power function with different bases&multiplicative factors yielding a non-zero answer.

Therefore, I think you just have to plug different k-values into your expression until you get the right answer.

Dumb exercise.
 
That's exactly why this question confused me. My work didn't turn out pretty, then I psyched myself out by thinking I must have read the question wrong. Thanks man.
 
One of the things that is confusing here, just as it was when you posted this same question before, is your use of the phrase "kth term of the series" when you started by giving a formula for the kth partial sum (of a geometric series). As I told you before, the "kth term of the series" [itex]\Sigma a_n[/itex] is ak. The sum a1+ a2+ ... + ak is the "kth partial sum".

You are given two geometric series, [itex]\Sigma a_n[/itex] and [itex]\Sigma b_n[/itex] . Do you want to find k so that the difference of the kth terms, ak- bk, is greater than 14, or so that the difference of the kth partial sums, [itex]\Sigma_{i=1}^k (a_n- b_n)[/itex] is greater than 14?
 
Last edited by a moderator:
Grrr... Argh... It's a retarded question.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K