Turns out to be an Integer

Algebra Level 3

$\large \dfrac{({1}+{2}+\cdots+{2017})({1}^2+{2}^2+\cdots+{2017}^2)}{{1}^3+{2}^3+\cdots+{2017}^3} = \quad ?$

The answer is 1345.

1 solution

Ravneet Singh
Jul 9, 2017

Let's consider generalized form:

$\large \dfrac{\displaystyle \sum_{k=1}^n k \times \displaystyle \sum_{k=1}^n k^2}{\displaystyle \sum_{k=1}^n k^3} = \dfrac{\left(\dfrac{n(n+1)}{2}\right){\left(\dfrac{n(n+1)(2n+1)}{6}\right)}}{\left(\dfrac{n(n+1)}{2}\right)^2} = \dfrac{2n+1}{3}$

For $n=2017$ , expression equals $\large \dfrac{4035}{3} = \boxed{1345}$

Turns out to be an Integer

The answer is 1345.

1 solution

0 pending reports