Algebra Challenge: Prove $\dfrac{2007}{2}-\ldots=\dfrac{2007}{2008}$

In summary, the Algebra Challenge is a test designed to strengthen understanding of algebraic concepts, such as solving equations and manipulating fractions. It helps improve problem-solving skills by requiring critical thinking and logical reasoning. The significance of proving the equation $\dfrac{2007}{2}-\ldots=\dfrac{2007}{2008}$ lies in solidifying the understanding of fraction manipulation and the concept of infinite series. It is necessary to show all the steps in the proof to identify errors and demonstrate a clear understanding of the process. To approach solving the Algebra Challenge, one should simplify each side of the equation, look for patterns, and use algebraic techniques such as factoring, combining like terms, and finding common denominators.
  • #1
anemone
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Prove that

$\dfrac{2007}{2}-\dfrac{2006}{3}+\dfrac{2005}{4}-\cdots-\dfrac{2}{2007}+\dfrac{1}{2008}=\dfrac{1}{1005}+\dfrac{3}{1006}+\dfrac{5}{1007}+\cdots+\dfrac{2007}{2008}$
 
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  • #2
Solution proposed by other that I wanted to share it here:

Let the LHS expression be $P$ and the RHS expression be $Q$.

$R=\left(\dfrac{2007}{2}+\dfrac{2006}{3}+\dfrac{2005}{4}+\cdots+\dfrac{1005}{1004}\right)+\left(\dfrac{1004}{1005}+\dfrac{1003}{1006}+\cdots+\dfrac{2}{2007}+\dfrac{1}{2008}\right)$

to both sides, and show that $P+R=Q+R$.

$\begin{align*} P+R&=\left(\dfrac{1}{1005}+\dfrac{1004}{1005}\right)+\left(\dfrac{3}{1006}+\dfrac{1003}{1006}\right)+\cdots+\left(\dfrac{2007}{2008}+\dfrac{1}{2008}\right)+\left(\dfrac{2007}{2}+\dfrac{2006}{3}+\cdots+\dfrac{1005}{1004}\right)\\&=1004+\left(\dfrac{2007}{2}+\dfrac{2006}{3}+\cdots+\dfrac{1005}{1004}\right)\end{align*}$

$\begin{align*} Q+R&=2\left(\dfrac{2007}{2}+\dfrac{2005}{4}+\cdots+\dfrac{1}{2008}\right)\\&=2007+\left(\dfrac{2005}{2}+\dfrac{2003}{3}+\cdots+\dfrac{3}{1003}+\dfrac{1}{1004}\right)\end{align*}$

Subtracting the above from below gives:

$\begin{align*} Q-P&=(2007-1004)+\left(\dfrac{2005-2007}{2}\right)+\left(\dfrac{2003-2006}{3}\right)+\cdots+\left(\dfrac{3-1006}{1003}+\right)+\left(\dfrac{1-1005}{1004}\right)\\&=1003+(-1003)\\&=0\end{align*}$

or simply $P=Q$ and hence we're done.
 

FAQ: Algebra Challenge: Prove $\dfrac{2007}{2}-\ldots=\dfrac{2007}{2008}$

What is the purpose of the Algebra Challenge?

The purpose of the Algebra Challenge is to test and strengthen one's understanding of algebraic concepts, such as solving equations and manipulating fractions.

How does the Algebra Challenge help improve problem-solving skills?

The Algebra Challenge requires critical thinking and logical reasoning to come up with a solution. By practicing these skills, one can improve their problem-solving abilities not just in algebra, but in other areas as well.

What is the significance of proving $\dfrac{2007}{2}-\ldots=\dfrac{2007}{2008}$?

This equation represents a common algebraic pattern and proving it helps solidify the understanding of fraction manipulation and the concept of infinite series.

Is it necessary to show all the steps in the proof?

Yes, showing all the steps in the proof is important as it allows one to identify any errors and demonstrate a clear understanding of the process. It also helps others follow and understand the solution.

How can I approach solving the Algebra Challenge?

Start by simplifying each side of the equation and looking for patterns. Then, use algebraic techniques such as factoring, combining like terms, and finding common denominators to manipulate the equation and prove its validity.

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