16.14

Algebra Level 4

n = 0 2016 n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) n 9 + 1 = ? \large \sum_{n=0}^{2016}\left \lfloor \frac{n(n+1)(n+2)(n+3)(n+4)}{n^9 +1}\right \rfloor = \ ?

Notation: \lfloor \cdot \rfloor denotes the floor function .


The answer is 61.

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2 solutions

n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) n 9 + 1 = { 60 n = 1 1 n = 2 0 n < 1 , n > 2 \left \lfloor \frac{n(n+1)(n+2)(n+3)(n+4)}{n^9 +1}\right \rfloor=\begin{cases}60 & n=1\\1 & n=2\\0&n<1,n>2\end{cases}

Hence, the sum is simply 60 + 1 = 61 60+1=61 .

n = 0 2016 n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) n 9 + 1 = n = 0 2 n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) n 9 + 1 + n = 3 2016 n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) n 9 + 1 = 0 1 + 5 ! 1 9 + 1 + 6 ! 2 2 9 + 1 + n = 3 2016 n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) n 9 + 1 = 0 + 60 + 1.4035 + s u m m a t i o n m u c h l e s s t h a n . 5 . = 61. \displaystyle \large \sum_{n=0}^{2016}\left \lfloor \frac{n(n+1)(n+2)(n+3)(n+4)}{n^9 +1}\right \rfloor \\ = \displaystyle \large \sum_{n=0}^2\left \lfloor \frac{n(n+1)(n+2)(n+3)(n+4)}{n^9 +1}+ \sum_{n=3}^{2016} \frac{n(n+1)(n+2)(n+3)(n+4)}{n^9 +1}\right \rfloor \\ = \displaystyle \left \lfloor \dfrac 0 1+ \dfrac {5!} {1^9+1}+ \dfrac{\frac{6!} 2}{ 2^9+ 1} + \sum_{n=3}^{2016} \frac{n(n+1)(n+2)(n+3)(n+4)}{n^9 +1}\right \rfloor \\ =\left \lfloor 0+60+1.4035~+~ summation~much~ less~than~.5 \right \rfloor.\\ = \Large \color{#D61F06}{61}.

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