How to find k term in a series. I've a series in which the nth term repeats n times.
For example see this :
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ...  
So the question which I want to ask is that is there any way (like a formula) to get the kth term from the series  (say 2000th term) except direct finding each and every term.
I also request to tell the derivation.
Note: This is not my HW question  (although it may look like! ), I rather prefer doing them by myself. 
 A: $m-$th term of the sequence is equal to $n$ if and only if the inequality:
$$1+2+...+(n-1)+1\le m \le 1+2+...+(n-1)+n \Leftrightarrow$$
$$\Leftrightarrow \frac{(n-1)n}{2}+1\le m\le\frac{n(n+1)}{2} \Leftrightarrow$$
$$\Leftrightarrow n^2-n+2 \le 2m \le n^2+n \Leftrightarrow$$
$$\Leftrightarrow n^2- n+\frac14< m < n^2+n+\frac14 \Leftrightarrow$$
$$\Leftrightarrow n-\frac12 < \sqrt{2m} < n+\frac12 \Leftrightarrow$$
$$\Leftrightarrow n<\sqrt{2m}+\frac12<n+1$$
$$n=\lfloor\sqrt{2m}+\frac12 \rfloor$$
Answer:
$$a_m=\lfloor\sqrt{2m}+\frac12 \rfloor, m\ge 1$$
A: Rather than try directly find the number associated with a position in the sequence, let us instead first try to find the position in the sequence associated with a number.  That will turn out to be easier, and we can then use that information as a kind of "bracket" to identify the number associated with a given position.
Specifically, how far do the $1$'s extend?  Well, there's only one $1$, and it's in the first position.
How far do the $2$'s extend?  There are two $2$'s, and they begin immediately after the $1$, so the last one is in the third ($1+2 = 3$) position.
How far do the $3$'s extend?  There are three $3$'s, and they begin immediately after the $2$'s, so the last one is in the sixth ($1+2+3 = 6$) position.
Hopefully, you will now see the pattern.  If you want to find the last number $n$, whatever $n$ happens to be, you will find it in the $1+2+3+\cdots+n$ position.  Numbers of this form are called triangular numbers:

In that article, you will also find a formula for computing the $n$th triangular number directly, rather than summing up the first $n$ integers:
$$
T_n = \frac{n(n+1)}{2}
$$
This formula is key to the process of finding the number at a given position.  We can say, for instance, that the number at a given position $k$ is $4$ if $k$ is in the fourth row of the triangle—that is, if we write the series $a_1, a_2, a_3, \ldots$, then $a_k = 4$ if and only if
$$
T_3 < k \leq T_4
$$
That is, the number cannot be in the first three rows (the triangle represented by $T_3$), and it must be in the first fourth rows (the triangle represented by $T_4$).  Ergo, it must be in the fourth row.
In much the same way, in general, $a_k = n$ if and only if
$$
T_{n-1} < k \leq T_n
$$
Using our formula above, we have
$$
\frac{(n-1)n}{2} < k \leq \frac{n(n+1)}{2}
$$
This gives us a condition on $k$ in terms of $n$; we want to flip that around to give a condition on $n$ in terms of $k$.  That way, we will have a formula that will return the value of $n = a_k$.  First multiply through by $2$:
$$
(n-1)n < 2k \leq n(n+1)
$$
$$
n^2-n < 2k \leq n^2+n
$$
We do something called "completing the square": We add something to the first and last expressions in order to put them in the form of a perfect square.  (Of course, we'll have to add the same thing to the middle expression as well).
$$
n^2-n+\frac{1}{4} < 2k+\frac{1}{4} \leq n^2+n+\frac{1}{4}
$$
Because we've completed the square, we can write this as
$$
\left(n-\frac{1}{2}\right)^2 < 2k+\frac{1}{4} \leq \left(n+\frac{1}{2}\right)^2
$$
We then take the square root of all sides:
$$
n-\frac{1}{2} < \sqrt{2k+\frac{1}{4}} \leq n+\frac{1}{2}
$$
We subtract $1/2$ from all sides to obtain
$$
n-1 < \sqrt{2k+\frac{1}{4}}-\frac{1}{2} \leq n
$$
In other words, $n$ must be the smallest integer greater than or equal to the middle expression.  This can be written more succinctly as
$$
n = \left\lceil \sqrt{2k+\frac{1}{4}}-\frac{1}{2} \right\rceil
$$
