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?

Numerator :
\begin{eqnarray*}
T_n &=& 1+3+5+...+(2n-1)\\
&=& 2+4+6+...+2n - n\\
&=& 2(1+2+3+...+n) - n\\
&=& 2\frac{n(n+1)}{2} - n\\
&=& n^2
\end{eqnarray*}
Similarly for denominator (check) :
\begin{eqnarray*}
B_n &=& 2+4+6+...+2n\\
&=& n(n+1)
\end{eqnarray*}
But it back in the initial equation and you get :
$$\frac{n^2}{n(n+1)} = \frac{2011}{2012} \Longrightarrow 2012n = 2011(n+1)$$
Last line is easy enough to solve.
So in fact the fraction is written in the form $$\frac{4044121}{4046132} = \frac{2011}{2012}$$
I hope this explains well enough why assuming the sums are $2011$ and $2012$ is wrong.
