Find the $2015$th Term in Sequence $1,2,-2,3,-3,3,4,-4,4,-4,5,-5,5,-5,5\cdots$

  • MHB
  • Thread starter anemone
  • Start date
  • Tags
    2015
In summary, the given sequence alternates between positive and negative integers, with each integer increasing by 1 until it reaches a certain point where it repeats the same integer before continuing the pattern. To find the $2015$th term, a formula of $a_n = a_1 + (n-1)d$ can be used, where $a_1=1$ and $d=1$ for the first 4 terms, and then the common difference changes to -1. There is a general formula for finding the $n$th term, which is $a_n = a_1 + (n-1)d$, with $a_1=1$ and alternating values of $d$ depending on the term
  • #1
anemone
Gold Member
MHB
POTW Director
3,883
115
Find the $2015$th term in the sequence $1,\,2,\,-2,\,3,\,-3,\,3,\,4,\,-4,\,4,\,-4,\,5,\,-5,\,5,\,-5,\,5\cdots$.

_______________________________________________________________________________________________________

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
  • #2
Congratulations to the following members for their correct solutions::)

1. kaliprasad
2.
greg1313
3. lfdahl
4. MarkFL

Solution from MarkFL:

To determine the $n$th term $a_n$ of this sequence, it is obvious we must make use of triangular numbers $T(m)$, where:

\(\displaystyle T(n)\equiv\frac{m(m+1)}{2}\)

We see that for:

\(\displaystyle n\in\left[T(m)-(m-1),T(m)\right]=\left[\frac{m^2-m+2}{2},\frac{m(m+1)}{2}\right]\)

we must have:

\(\displaystyle \left|a_n\right|=m\)

We can then observe that we must have:

\(\displaystyle a_n=(-1)^{\frac{2n-m^2+m-2}{2}}m\)

where:

\(\displaystyle m^2-m+2\le2n\le m^2+m\)

For $n=2015$, we then find:

\(\displaystyle 63^2-63+2\le2\cdot2015\le63^2+63\)

\(\displaystyle 3908\le4030\le4032\)

Hence, for $n=2015$, we have $m=63$, and so:

\(\displaystyle a_{2015}=(-1)^{\frac{2\cdot2015-63^2+63-2}{2}}\cdot63=(-1)^{61}63=-63\)
 

FAQ: Find the $2015$th Term in Sequence $1,2,-2,3,-3,3,4,-4,4,-4,5,-5,5,-5,5\cdots$

What is the pattern of this sequence?

The pattern of this sequence is that it alternates between positive and negative integers, with each integer increasing by 1 until it reaches a certain point where it repeats the same integer before continuing the pattern.

How can I find the $2015$th term in this sequence?

To find the $2015$th term in this sequence, you can use the formula $a_n = a_1 + (n-1)d$, where $a_1$ is the first term in the sequence, $n$ is the term number, and $d$ is the common difference. In this sequence, $a_1=1$, $d=1$ for the first 4 terms, and then the common difference changes to -1. So the formula would be $a_{2015}=1+(2015-1)(-1)=1-2014=-2013$.

Is there a general formula for finding the $n$th term in this sequence?

Yes, the general formula for finding the $n$th term in this sequence is $a_n = a_1 + (n-1)d$, where $a_1$ is the first term in the sequence, $n$ is the term number, and $d$ is the common difference. In this sequence, $a_1=1$, and $d$ alternates between 1 and -1, depending on the term number.

What is the significance of the repeating terms in this sequence?

The repeating terms in this sequence serve as a break in the pattern, indicating that the common difference has changed. This break occurs every 4 terms, with the first 4 terms having a common difference of 1, and the next 4 terms having a common difference of -1.

Can this sequence be extended to include more terms?

Yes, this sequence can be extended to include more terms, as there is a clear pattern and a general formula for finding the $n$th term. As long as the common difference continues to alternate between positive and negative integers, the sequence can be extended indefinitely.

Back
Top