Find the value of the sum below

Algebra Level 3

$\dfrac11 + \dfrac1{1 + 2} + \dfrac1{1 + 2 + 3} + \dfrac1{1 + 2 + 3 + 4} + \cdots+ \dfrac1{1 + 2 + 3 + \cdots+ 2017}$

If the sum above simplifies to $\dfrac ab$ , where $a$ and $b$ are coprime positive integers, then find $a + b$ .

1 solution

Md Zuhair
Mar 21, 2017

$\large \displaystyle = \sum_{n=1}^{2017} \frac{1}{\frac{n(n+1)}{2}}$

$\large \displaystyle = 2\sum_{n=1}^{2017} \frac{1}{n(n+1)}$

$\large \displaystyle = 2\sum_{n=1}^{2017} \frac{1}{n} - \frac{1}{n+1}$

$\large \displaystyle = 2 \Big (1 - \frac{1}{2018} \Big)$

$\large \displaystyle = \frac{2017}{1009}$

Making the answer $\color{#3D99F6} \boxed{ \large \displaystyle 3026 }$ .

My solution from the reports! Thank you! :)

Guilherme Niedu - 4 years, 2 months ago

Identified well

Md Zuhair - 4 years, 2 months ago