Solve 6th Grade Sequence: 1,5,13,25,41,61

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In summary, The conversation is about a parent asking for help with their child's math homework. The problem is to find an equation for the sequence 1, 5, 13, 25, 41, 61. The conversation includes hints and ideas for solving the problem, with the final equation being a_{n}=n^{2}+(n-1)^{2}.
  • #1
debrawallenger
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Help, anyone. I am very new to this forum and am really here trying to help my son with his wicked 6th grade homework. His teacher is in my opinion, assigning problems that are way too difficult. I was hoping someone here could maybe help. The problem is as follows: The students are given the sequence 1,5,13,25,41,61 and have to come up with an equation to solve the sequence. Any ideas? This should be easy for you all. But for me, who was good at math at one time, this is beyond what I can come up with. Any help would be appreciated!
 
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  • #2
Hint: what's the difference between 5 and 1? 13 and 5? 25 and 13? 41 and 25? 61 and 41?
see the pattern?

oh, and Hi! Welcome to PF! :smile:
 
  • #3
Hmm, I see a pattern too. Counting up from zero 1, 4, 8, 12, 16, 20...
 
  • #4
I see the pattern as well, they are all separated by multiples of four, but it's coming up with the equation that is the problem. Any more hints? Maybe it will job something in my distance math past!
 
  • #5
Sorry about that, I bungled up a bit..
 
  • #6
My mistake everyone,

I am going to post in the homework section. Thanks all. I also stated the problem incorrectly.

debra
 
  • #7
From what I can see, the terms fulfill:
[tex]a_{n}=n^{2}+(n-1)^{2}[/tex]
 

FAQ: Solve 6th Grade Sequence: 1,5,13,25,41,61

What is the pattern in the sequence 1,5,13,25,41,61?

The pattern in this sequence is that each term is obtained by adding a multiple of 4 to the previous term. Specifically, the first term is 1, and then each subsequent term is obtained by adding 4 times the position of the term in the sequence minus 3. For example, the second term is obtained by adding 4*(2-1) = 4 to the first term, resulting in 5. The third term is obtained by adding 4*(3-1) = 8 to the second term, resulting in 13, and so on.

What is the sixth term in the sequence 1,5,13,25,41,61?

The sixth term in the sequence is 61.

What is the rule for finding the nth term in the sequence 1,5,13,25,41,61?

The rule for finding the nth term in this sequence is n^2 + 1. This means that to find the nth term, you square the position of the term in the sequence and then add 1. For example, the fourth term is obtained by squaring 4 and then adding 1, resulting in 17 (4^2 + 1 = 17).

How can this sequence be extended beyond the given terms?

This sequence can be extended by continuing the pattern of adding 4*(n-1) to the previous term. For example, the seventh term in the sequence would be obtained by adding 4*(6-1) = 20 to the sixth term, resulting in 81 (61 + 20 = 81).

What is the relationship between this sequence and the Fibonacci sequence?

There is no direct relationship between this sequence and the Fibonacci sequence. However, both sequences can be generated using a recursive formula, and both involve adding previous terms to generate new terms. Additionally, both sequences have values that increase at an increasing rate.

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