Simple Simplification II

$\left\lfloor\frac{((2017 \times\ 2016 \times\ 2015)+1)2014}{((2016 \times\ 2015) + 2015)2014}\right\rfloor \ = \left\lfloor\frac{((2017 \times\ 2016 \times\ 2015)+1)}{(2017 \times\ 2015) } \right\rfloor$ $= \left\lfloor \ 2016 + \frac{1}{2017 \times 2015}\right\rfloor = \boxed{2016}$

Let $n = 2016$

$\dfrac{(n+1)! + (n-2)!}{n!+(n-1)!}$

$= \dfrac{(n+1)n(n-1)+1}{n(n-1)+(n-1)}$

$= \dfrac{n(n^2-1)+1}{n^2-1} = n+\dfrac 1{n^2-1}$

Then the answer is $2016$

• We can rewrite the expression as $\frac{2014!(2017\times2016\times2015 + 1)}{2015!(2016 + 1)} = \frac{2017\times2016\times2015 + 1}{2015\times2017}$ .
• We can expand this expression as $\frac{2017\times2016\times2015}{2017\times2015} + \frac{1}{2017\times2015}$ and it equals $2016 + \frac{1}{2017\times2015}$ ,
• Since we want the greatest integer less than or equal to x and the other fraction is obviously smaller than 1., the answer comes up as 2016 .

[(2014!(2017x2016x2015+1/(2015!(2016+1))] =[(2017x2016x2015+1)/2015x2017 =[2016+1/(2015x2016)] which is very close to 2016

A four line java program can be used to do it much faster. Can anyone please suggest any method to do the sum mentally ?