2017⁴ⁿ có tận cùng là 1 ; 2015ⁿ có tận cùng là 5.

Ta có: A = 2017²⁰¹⁶-2015²⁰¹⁴ = 2017^4.504-2015²⁰¹⁴ = (...1)-(...5) = (...6)

A có chữ số tận cùng là 6 nên khi chia A cho 5 sẽ dư 1

=(2014/2014)+(2015+2015)+(2016/2016)+(2017+2017)

=1+1+1+1

=4

vậy A=4 (4=4)

Dấu < nhé!

2014+2015+2016/2015+2016+2017<2014/2015+2015/2016+2016/2017

dấu = đấy

(2016/2017) = (2017/2016)

\(A=\dfrac{2014}{2015}+\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2014}\\ =1-\dfrac{1}{2015}+1-\dfrac{1}{2016}+1-\dfrac{1}{2017}+1+\dfrac{1}{2014}+\dfrac{1}{2014}+\dfrac{1}{2014}\\ =\left(1+1+1+1\right)+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]\\ =4+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]\)

Vì \(\dfrac{1}{2015}< \dfrac{1}{2014}\), \(\dfrac{1}{2016}< \dfrac{1}{2014}\), \(\dfrac{1}{2017}< \dfrac{1}{2014}\)

\(\Rightarrow\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)< 0\\ \Rightarrow-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\\>0\\ \Rightarrow4+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]>4\)

A = (n + 2015)(n + 2016) + n² + n

= (n + 2015)(n + 2015 + 1) + n(n + 1)

Tích 2 số tự nhiên liên tiếp luôn chia hết cho 2

=> (n + 2015)(n + 2015 + 1) chia hết cho 2

      n(n + 1) chia hết cho 2

=> (n + 2015)(n + 2015 + 1) + n(n + 1) chia hết cho 2

=> A chia hết cho 2 với mọi n \(\in\) N (đpcm)