Evaluate
$$2004\cdot\left(\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4}+\dots + \frac{1}{2003\cdot2004}\right)$$
I think that this is based on either geometric or arithmetic progressions but I am not sure.
Please help me solve this problem.
Answer
You should know that $\dfrac{1}{x(x+1)}=\dfrac{1}{x}-\dfrac{1}{x+1}$.
Apply this to your series. I will ignore the $2004$ factor for now.
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots+\frac{1}{2003\cdot 2004}=1\color{red}{\underbrace{-\frac{1}{2}+\frac{1}{2}}}\color{#AB4DEF}{\underbrace{-\frac{1}{3}+\frac{1}{3}}}\dots\color{orange}{\underbrace{-\frac{1}{2003}+\frac{1}{2003}}}-\frac{1}{2004}$$
See how you can cancel pretty much all the terms? What you are left with is $1-\dfrac{1}{2004}$, which equals $\dfrac{2003}{2004}$. Now don't forget that $2004$ factor at the left.
$$\color{blue}{2004}\cdot\frac{2003}{\color{blue}{2004}}=2003$$
$$\displaystyle \color{green}{\boxed{\therefore 2004\cdot\left(\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+\dots + \dfrac{1}{2003\cdot 2004}\right)=2003}}$$
If you want to see why $\dfrac{1}{x(x+1)}=\dfrac{1}{x}-\dfrac{1}{x+1}$, it is easy to prove. Just use LHS-RHS. I will try to make $\dfrac{1}{x}-\dfrac{1}{x+1}$ equal to $\dfrac{1}{x(x+1)}$.
$$\frac{1}{x}-\dfrac{1}{x+1}=\frac{x+1}{x(x+1)}-\frac{x}{x(x+1)}=\frac{x+1-x}{x(x+1)}=\frac{1}{x(x+1)}$$
$$\therefore \frac{1}{x}-\frac{1}{x+1}=\frac{1}{x(x+1)} \ \ \text{by LHS-RHS}$$
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