Welcome 2016! Part 20

Algebra Level 4

1 1 + 2 + 1 1 + 2 + 3 + ⋯ + 1 1 + 2 + ⋯ + 20 \large \dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\cdots+\dfrac{1}{1+2+\cdots+20}

If the value of the expression above is equal to m n \dfrac mn , where m m and n n are coprime positive integers, find the value of m + n m+n .


The answer is 40.

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Anish Harsha by Anish Harsha , 20, India

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2 solutions

Akhil Bansal
Dec 26, 2015

Given expression equals,
= ∑ i = 2 20 1 ∑ n = 1 i n \large = \displaystyle \sum_{i=2}^{20} \dfrac{1}{\sum_{n=1}^i n } = ∑ i = 2 20 2 i ( i + 1 ) \large = \sum_{i=2}^{20} \dfrac{2}{i(i+1)} It's a telescoping series , = 2 [ ∑ i = 2 20 ( 1 i − 1 i + 1 ) ] \large = 2\left[ \displaystyle \sum_{i=2}^{20}\left( \dfrac{1}{i} - \dfrac{1}{i+1}\right) \right] = 2 [ 1 2 − 1 3 + 1 3 − … − 1 21 ] \large = 2\left[ \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \ldots -\dfrac{1}{21}\right] ⇒ 2 [ 1 2 − 1 21 ] = 19 21 \large \Rightarrow 2\left[\dfrac{1}{2} - \dfrac{1}{21}\right] = \color{#3D99F6}{ \dfrac{19}{21}}

Note : Sum of n consecutive natural numbers = n ( n + 1 ) 2 \dfrac{n(n+1)}{2}

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Moderator note:

Good observation of the telescoping series.

Going a step further, if you take the upper limit to be x x instead of 20 20 , you can generalize it:

2 ∑ n = 2 x 1 n − 1 n + 1 = x − 1 x + 1 , x ≥ 2 2\sum _{ n=2 }^{ x }{ \frac { 1 }{ n } -\frac { 1 }{ n+1 } =\frac { x-1 }{ x+1 } ,\quad x\ge 2 }

Andrew Tawfeek - 5 years, 1 month ago
Rohit Udaiwal
Dec 26, 2015

First recall that sum of fisrt n n natural numbers is given by n ( n + 1 ) 2 \dfrac{n(n+1)}{2} .So sum of 1 + 2 = 2 ( 2 + 1 ) 2 , 1 + 2 + 3 = 3 ( 3 + 1 ) 2 \color{#20A900}{1+2=\dfrac{2(2+1)}{2}},\color{#3D99F6}{1+2+3=\dfrac{3(3+1)}{2}} and so on.Our expression then becomes: 1 2 ⋅ 3 2 + 1 3 ⋅ 4 2 + … + 1 20 ⋅ 21 2 = 2 2 ⋅ 3 + 2 3 ⋅ 4 + … + 2 20 ⋅ 21 = 2 ( 1 2 ⋅ 3 + 1 3 ⋅ 4 + … + 1 20 ⋅ 21 ) = 2 ( 1 2 − 1 3 + 1 3 − 1 4 + … + 1 20 − 1 21 ) = 2 ( 1 2 − 1 21 ) = 2 ⋅ 19 42 = 19 21 \dfrac{1}{\color{#20A900}{\frac{2\cdot3}{2}}}+\dfrac{1}{\color{#3D99F6}{\frac{3\cdot4}{2}}}+\ldots+\dfrac{1}{\color{magenta}{\frac{20\cdot21}{2}}}=\color{#20A900}{\dfrac{2}{2\cdot3}}+\color{#3D99F6}{\dfrac{2}{3\cdot4}}+\ldots+\color{magenta}{\dfrac{2}{20\cdot21}} \\ =2\left(\dfrac{1}{\color{#20A900}{2\cdot3}}+\dfrac{1}{\color{#3D99F6}{3\cdot4}}+\ldots+\dfrac{1}{\color{magenta}{20\cdot21}}\right) \\ =2\left(\color{#20A900}{\dfrac{1}{2}}-\color{#20A900}{\dfrac{1}{3}}+\color{#3D99F6}{\dfrac{1}{3}}-\color{#3D99F6}{\dfrac{1}{4}}+\ldots+\color{magenta}{\dfrac{1}{20}}-\color{magenta}{\dfrac{1}{21}}\right) \\ =2\left(\dfrac{1}{2}-\dfrac{1}{21}\right)=2\cdot\dfrac{19}{42} \\ =\boxed{\dfrac{19}{21}}

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