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

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Find the $2015$th term in the sequence $1,\,2,\,-2,\,3,\,-3,\,3,\,4,\,-4,\,4,\,-4,\,5,\,-5,\,5,\,-5,\,5\cdots$.

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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:

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