16.13

Algebra Level 4

$\large \sum_{n=0}^{2016} \left \lfloor \dfrac{ n^3 + n^2+ 1}{n^2 + n+1} \right \rfloor = \, ?$

The answer is 2031122.

1 solution

A Former Brilliant Member
Jun 13, 2016

$\sum_{n=0}^{2016} \left \lfloor \dfrac{ n^3 + n^2+ 1}{n^2 + n+1} \right \rfloor = \sum_{n=0}^{2016} \left \lfloor \dfrac{n^3-1 +n^2+2}{n^2 + n+1} \right \rfloor\\ =\sum_{n=0}^{2016} \left \lfloor n-1+ \dfrac{ n^2+ 2}{n^2 + n+1} \right \rfloor\\\dfrac{n^2+2}{n^2+n+1}=\begin{cases}2 & n = 0\\1 & n = 1\\0&n>1\end{cases}\\\therefore\sum_{n=0}^{2016} \left \lfloor n-1+ \dfrac{ n^2+ 2}{n^2 + n+1} \right \rfloor=\sum_{n=0}^{2016}n-2017+3=2031122$

16.13

The answer is 2031122.

1 solution

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