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

  • Context: High School 
  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    2015
Click For Summary
SUMMARY

The $2015$th term in the sequence $1, 2, -2, 3, -3, 3, 4, -4, 4, -4, 5, -5, 5, -5, 5\cdots$ can be determined by analyzing the pattern of the sequence. The sequence alternates between positive and negative integers, with repetitions of certain values. The correct term was identified by members kaliprasad, greg1313, lfdahl, and MarkFL, with MarkFL providing the final solution that confirmed the term's position accurately.

PREREQUISITES
  • Understanding of integer sequences
  • Familiarity with pattern recognition in mathematics
  • Basic knowledge of mathematical notation
  • Ability to analyze sequences for repetition and alternation
NEXT STEPS
  • Study integer sequence generation techniques
  • Learn about mathematical induction for proving sequence properties
  • Explore combinatorial methods for sequence analysis
  • Investigate the use of generating functions in sequences
USEFUL FOR

Mathematicians, educators, students studying sequences, and anyone interested in advanced pattern recognition in mathematics.

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
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
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:

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

We see that for:

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

we must have:

$$\left|a_n\right|=m$$

We can then observe that we must have:

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

where:

$$m^2-m+2\le2n\le m^2+m$$

For $n=2015$, we then find:

$$63^2-63+2\le2\cdot2015\le63^2+63$$

$$3908\le4030\le4032$$

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

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

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K