How Can You Solve a Geometric Sequence Problem in a Telephoning Tree?

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To solve the geometric sequence problem in a telephoning tree, the correct formula to determine the level at which 64 students are contacted is 2^n - 1. The calculations show that at level 6, 62 students are contacted, while level 7 reaches 64 students. The confusion arises from the interpretation of the nth level, which refers to the total number of students contacted up to that level. Therefore, the answer for part a) is that 64 students are contacted at the 7th level, and for part c), the nth level indicates the cumulative total of students contacted. Understanding the geometric sequence is crucial for accurately determining these values.
JDK
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Hi,

I have a relatively simple question. In this particular problem in my Math 30 Pure textbook...

10. Here are three levels in a school telephoning tree.

Teacher
Student 1 Student 2
Student 3 Student 4 Student 5 Student 6

a)At what level are 64 students contacted?

...I really need to know if there is a formula I can use to get the answer. I've tried using tn = (a)(r^(n-1)), and Sn = a((r^n)-1)/r-1 while replacing both tn and Sn with 64 and solving. Either way I end up with incorrect answers. The correct response, as said by the book is 7 or the 7th level. I also tried taking this line from my calculations... Sn = 1((2^n)-1) and graphed it as y = 1((2^x)-1). I get the correct values for 'y' (# of students) but the x values are one less then they should be in each case. For example, at x=6, y=63 (one off from the value stated in the question). The answer is so close to being correct but the level is off by one as well as the 'y' value. Is there an explanation to this or something I missed about gemoetric sequences. Any help would be awesome! Thanks a million.

Also, part c) of the same question asks - 'By the nth level how many students have been contacted? What exactly is the nth level?

- JDK
 
Mathematics news on Phys.org
Look in Level1 : There is one Element (teacher)
Level2 : There are 2 Elements = 22-1 (student)
Level3 : There are 4 Elements = 23-1 (student)
...
...
Leveln : There are 2n-1 elements

So total no of elements upto nth level will be
1+2+4+...2n-1= 2n-1

for part c)
now students contacted by nth level
would be 2n-2 subtracting one teacher

For part a)

2n-2=64 when
n=6.044
Now n is an integer so n=6 OR 7

But for n=6 26-2=62 students would have been contacted so
n=7 is the level when 64th student have been contacted
 
Good morning I have been refreshing my memory about Leibniz differentiation of integrals and found some useful videos from digital-university.org on YouTube. Although the audio quality is poor and the speaker proceeds a bit slowly, the explanations and processes are clear. However, it seems that one video in the Leibniz rule series is missing. While the videos are still present on YouTube, the referring website no longer exists but is preserved on the internet archive...

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