determining $n$ in a given sequence $\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $

Question

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?

Answer 1

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.

Answer 2

$1 + 3 + 5 + .. + 2n - 1 = n^2$, 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 $n = 2011$