Generalize for n

Algebra Level 3

1 1 × 3 + 1 3 × 5 + 1 5 × 7 + ⋯ + 1 2015 × 2017 = m n \large \dfrac{1}{1\times 3}+\dfrac{1}{3\times 5}+\dfrac{1}{5\times 7}+\dots+\dfrac{1}{2015\times 2017}=\dfrac{m}{n} If m m and n n are coprime integers, then find the value of m + n m+n .


The answer is 3025.

Áron Bán-Szabó by Áron Bán-Szabó , 17, Hungary

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Zach Abueg
Aug 29, 2017

S = 1 1 × 3 + 1 3 × 5 + 1 5 × 7 + ⋯ + 1 2015 × 2017 = ∑ n = 1 1008 1 ( 2 n − 1 ) ( 2 n + 1 ) = 1 2 ∑ n = 1 1008 ( 1 2 n − 1 − 1 2 n + 1 ) = 1 2 ( 1 − 1 2017 ) = 1008 2017 \displaystyle \begin{aligned} S & = \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \cdots + \frac{1}{2015 \times 2017} \\ & = \sum_{n \ = \ 1}^{1008} \frac{1}{(2n - 1)(2n + 1)} \\ & = \frac 12 \sum_{n \ = \ 1}^{1008} \left( \frac{1}{2n - 1} - \frac{1}{2n + 1} \right) \\ & = \frac 12 \left(1 - \frac{1}{2017} \right) \\ & = \frac{1008}{2017} \end{aligned}

⟹ m + n = 1008 + 2017 = 3025 \implies m + n = 1008 + 2017 = \boxed{3025}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...