Hidden Harmonies

Algebra Level 2

1 1 × 2 + 1 2 × 3 + 1 3 × 4 + . . . + 1 2016 × 2017 + 1 2017 × 2018 = x \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{2016\times2017}+\dfrac{1}{2017\times2018}=x

What is x x ?

You don't need to use sigma, though you can.

Other \text{Other} 2017 2018 \dfrac{2017}{2018} 2017 2017 2016 2017 \dfrac{2016}{2017}

Terry Yu by Terry Yu , 17, USA

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3 solutions

Terry Yu
Jun 14, 2017

If you noticed,

1 1 × 2 = 1 1 − 1 2 \dfrac{1}{1\times2}=\dfrac{1}{1}-\dfrac{1}{2}

1 2 × 3 = 1 2 − 1 3 \dfrac{1}{2\times3}=\dfrac{1}{2}-\dfrac{1}{3}

and so on and so forth until you get the equation

1 1 − 1 2 + 1 2 − 1 3 + 1 3 . . . − 1 2017 + 1 2017 − 1 2018 \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}...-\dfrac{1}{2017}+\dfrac{1}{2017}-\dfrac{1}{2018}

which can simplify to

1 1 − 1 2 + 1 2 − 1 3 + 1 3 . . . − 1 2017 + 1 2017 − 1 2018 = 1 − 1 2018 = 2017 2018 \dfrac{1}{1}- \xcancel{\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}...-\dfrac{1}{2017}+\dfrac{1}{2017}}-\dfrac{1}{2018}=1-\dfrac{1}{2018}=\color{#D61F06}\large\boxed{\dfrac{2017}{2018}}

9 Helpful 0 Interesting 0 Brilliant 1 Confused

We can generalize this as the sum ∑ k = 1 n ( 1 n − 1 n + 1 ) \displaystyle \sum_{k \ = \ 1}^{n} \Bigg(\frac{1}{n} - \frac{1}{n + 1}\Bigg) , where 1 n ( n + 1 ) = 1 n − 1 n + 1 \displaystyle \frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1} through partial fraction decomposition. A series where terms cancel as you showed is called a telescoping series .

Zach Abueg - 3 years, 12 months ago
Chew-Seong Cheong
Jun 20, 2017

x = 1 1 × 2 + 1 2 × 3 + 1 3 × 4 + ⋯ + 1 2016 × 2017 + 1 2017 × 2018 = 1 1 − 1 2 + 1 2 − 1 3 + 1 3 − 1 4 + ⋯ + 1 2016 − 1 2017 + 1 2017 − 1 2018 = 1 1 − 1 2 + 1 2 − 1 3 + 1 3 − 1 4 + 1 4 + ⋯ − 1 2017 + 1 2017 − 1 2018 = 1 − 1 2018 = 2017 2018 \begin{aligned} x & = {\color{#3D99F6} \frac 1{1\times 2}} +{\color{#D61F06} \frac 1{2\times 3}} + {\color{#3D99F6} \frac 1{3\times 4}} + \cdots + {\color{#3D99F6} \frac 1{2016\times 2017}} +{\color{#D61F06} \frac 1{2017\times 2018}} \\ & = {\color{#3D99F6} \frac 11 - \frac 12} +{\color{#D61F06} \frac 12 - \frac 13} + {\color{#3D99F6} \frac 13 - \frac 14} + \cdots + {\color{#3D99F6} \frac 1{2016} - \frac 1{2017}} +{\color{#D61F06} \frac 1{2017} - \frac 1{2018}} \\ & = \frac 11 \cancel{- \frac 12 + \frac 12} \cancel{- \frac 13 + \frac 13} \cancel{- \frac 14 + \frac 14}+ \cdots \cancel{- \frac 1{2017} + \frac 1{2017}} - \frac 1{2018} \\ & = 1 - \frac 1{2018} \\ & = \boxed{\dfrac {2017}{2018}} \end{aligned}

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Hana Wehbi
Jun 15, 2017

x = ∑ 1 2017 1 n ( n + 1 ) = ∑ 1 2017 ( 1 n − 1 n + 1 ) = 1 − 1 2018 = 2017 2018 x= \sum_1^{2017 }\frac{1}{n(n+1)} = \sum_1^{2017}(\frac{1}{n} - \frac{1}{n+1} ) = 1-\frac{1}{2018}= \frac{2017}{2018} , keeping in mind that many elements will be canceled.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

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