Final step: Telescoping sum

Calculus Level 3

3 1 ! + 2 ! + 3 ! + 4 2 ! + 3 ! + 4 ! + ⋯ + 2017 2015 ! + 2016 ! + 2017 ! = 1 m ! − 1 n ! \large \begin{aligned}\dfrac{3}{1!+2!+3!} + \dfrac{4}{2!+3!+4!} + \cdots +\dfrac{2017}{2015!+2016!+2017!} = \dfrac{1}{m!} - \frac{1}{n!}\end{aligned} If m m and n n are coprime integers then, find the value of m + n m+n .


Notation: ! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .


The answer is 2019.

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

Ashish Menon
Sep 13, 2017

∑ n = 1 2015 [ r + 2 r ! + ( r + 1 ) ! + ( r + 2 ) ! ] = ∑ n = 1 2015 [ r + 2 r ! ( 1 + r + 1 + ( r + 1 ) ( r + 2 ) ] = ∑ n = 1 2015 [ 1 r ! ( r + 2 ) ] = ∑ n = 1 2015 [ r + 1 ( r + 2 ) ! ] = ∑ n = 1 2015 [ 1 ( r + 1 ) ! − 1 ( r + 2 ) ! ] = 1 2 ! − 1 2017 ! \begin{aligned} \displaystyle \sum_{n=1}^{2015} \left [ \dfrac{r + 2}{r! + (r+1)! + (r+2)!} \right ] & = \displaystyle \sum_{n=1}^{2015} \left [ \dfrac{r + 2}{r!(1 + r + 1 + (r+1)(r+2)} \right ] \\ \\ & = \displaystyle \sum_{n=1}^{2015} \left [ \dfrac{1}{r!(r+2)} \right ] \\ \\ & = \displaystyle \sum_{n=1}^{2015} \left [ \dfrac{r + 1}{(r+2)!} \right ]\\ \\ & = \displaystyle \sum_{n=1}^{2015} \left [ \dfrac{1}{(r+1)!} - \dfrac {1}{(r+2)!} \right ]\\ \\ & = \dfrac {1}{2!} - \dfrac {1}{2017!} \end{aligned}

∴ m + n = 2 + 2017 = 2019 \therefore \text {m + n} \ = \ 2 + 2017 \ = \boxed{2019} .

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