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Given that: $$\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $$ Determine $n$.enter image description here

The memorandum says the answer is 2011 but how is that so? Where did I go wrong?

Jyrki Lahtonen
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Aspiring Mathlete
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  • What are $a$ and $d$? Also the numerator might not be $2011$ and the denominator $2012$. It could be $\frac{4022}{4024}$ for example. – user88595 Apr 23 '14 at 08:36

2 Answers2

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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.

user88595
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  • What did you do in the second line of your solution? – Aspiring Mathlete Apr 23 '14 at 08:46
  • I added $1$ to each of the element and there are $n$ of them so I subtracted by $n$ at the end so that the equality is still valid. Check for $n = 3$ and $4$ if you want to convince yourself. – user88595 Apr 23 '14 at 08:48
  • OK. I figured out what you did, but how did you decide to subtract by $n$? How did you know you had to do that and not by 1? – Aspiring Mathlete Apr 23 '14 at 08:57
  • The sum contains $n$ terms. If I add $1$ to each, that's like adding $n$. Imagine you have $n$ people and you decide to give $1$ coin to each, you'll need $n$ coins. Because I have an equality, I need to adjust it so that's why I subtracted by $n$ after. Effectively, I added $0 = n-n$ – user88595 Apr 23 '14 at 09:01
  • If you are still not convinced, you could check that link which explains why the sum of odd numbers is $n^2$. http://demo.activemath.org/ActiveMath2/search/show.cmd?id=mbase://LeAM_calculus/proofs_school/ex_induction – user88595 Apr 23 '14 at 09:03
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$1 + 3 + 5 + .. + 2n - 1 = n^2$, 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 $n = 2011$

DeepSea
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