determining $n$ in a given sequence $\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $ Given that: $$\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $$
Determine $n$.
The memorandum says the answer is 2011 but how is that so? Where did I go wrong?

$1 + 3 + 5 + .. + 2n - 1 = n^2$, and $2 + 4 + ... + 2n = n(n + 1) = n^2 + n$. So:
$\dfrac{n^2}{n^2 + n} = \dfrac{2011}{2012}$. So: $2012n^2 = 2011n^2 + 2011n$, and $n^2 = 2011n$. So $n = 2011$
