Solving $a_n$ in terms of $n$ Let $a_n=\frac {1+2+3\ldots n}{1+3+5\ldots+(2n-1)}$
$(1)$ Find $a_1,a_2,a_3$ and $a_{10}$
$(2)$ Find $a_n$ in terms of $n$. You may use the pattern you discovered or some formulas you know.
$(3)$ Find the smallest integer $n$ such that $$a_n-\frac {1}{2}\leq\frac {1}{2012}$$

I'm not too worried on solving $(1)$ because it's easy. My main worry is $(2)$ and $(3)$.
For $(2)$, I think that we can replace with the sums with there formulas. So the numerator would become $\frac {n(n+1)}{2}$ and the denominator would be $n^2$. But I would like to verify if that is the correct path to take.
For $(3)$, my only thought would to add $\frac {1}{2}$ to both sides and get $$a_n=\frac {1007}{2012}$$. But I don't know what to do afterward.

Any help is welcome!
 A: $$a_n=\frac{\frac{n(n+1)}{2}}{n^2}$$
$$a_n=\frac{(n+1)}{2n}=\frac{1}{2} +\frac{1}{2n}$$
$$a_n-\frac {1}{2}\leq\frac {1}{2012}$$
$$\implies \frac{1}{2n} \le \frac{1}{2012}$$
$$\implies n\ge 1006 $$
A: Hints: Use $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^n (2k-1) = 2\sum_{k=1}^n k -\sum_{k=1}^n 1 = n(n+1) - n = n^2$.
Combining these equations you obtain the result Aakash Kumar posted. 
A: The numerator is $\frac{n(n+1)}{2}$ as there are $n$ uniformly distributed numbers whose average is $\frac{1+n}{2}$, i.e. the average of the first and last term. 
You can also do what Gauss did.
Consider $1+2+3..+(n-2)+(n-1)+n$ for $n$ even.
We have $1+n=2+(n-1)=3+(n-2)=..$ and we have $\frac{n}{2}$ groups of $n+1$ for a sum of $\frac{n}{2}(n+1)$. Then you can use this formula to figure out the formula for when $n$ is odd (using the sum of the first $n-1$ positive integers). 
The denominator is $n^2$ see this picture:

Or again there are $n$ uniformly distributed numbers and the average is $\frac{1+2n-1}{2}=n$.
(2) follows and (3) follows with some more algebra.
