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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)
\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)
\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)
\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)
\(2A=2+3+4+5+6+...+2012+2013+2014\)
\(2A=\dfrac{\left(2+2014\right).2013}{2}\)
\(A=\dfrac{2016.2013}{4}=504.2013\)
\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)
\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)
\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)
\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)
\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)
\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)
\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)
\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta thấy:
\(\frac{2012}{2013}+\frac{2013}{2014}>\frac{2012}{2014}+\frac{2013}{2014}=\frac{2012+2013}{2014}>\frac{2012+2013}{2013+2014}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
C = 2012 . 2016
C = 2012 . (2014 + 2)
C = 2012 . 2014 + 2012 . 2
D = 2014 . 2014
D = (2012 + 2) . 2014
D = 2012 . 2014 + 2014 . 2
Vì 2012 . 2 < 2014 . 2 nên 2012 . 2014 + 2012 . 2 < 2012 . 2014 + 2014 . 2
=> C < D
C = 2012 * 2016 = [2014 - 2][2014 + 2] = 20142 + 4028 - 4028 - 22 = 20142 - 22 [có thể áp dụng HĐT cho nhanh]
D = 2014 * 2014 = 20142
20142 - 22 < 20142
<=> C < D
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 2016^2015 +1 / 2016^2014+1 < 2016^2015 + 1 + 2015 / 2016^2014 + 1 + 2015
= 2016^2015 + 2016 / 2016^2014 + 2016
= 2016(2016^2014 + 1 ) / 2016(2016^2013 +1)
= 2016^2014 + 1 / 2016^2013 + 1 = B
=> A < B
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét N có:
\(N=\frac{2012+2013+2014}{2013+2014+2015}=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)
Ta các số hạng của M và N có:
\(\frac{2012}{2013}>\frac{2012}{2013+2014+2015}\) (1)
\(\frac{2013}{2014}>\frac{2013}{2013+2014+2015}\) (2)
\(\frac{2014}{2015}>\frac{2014}{2013+2014+2015}\) (3)
Từ (1);(2);(3) => M > N
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
\(\frac{2012+2013.2014}{2014.2015-2016}\)
\(=\frac{2013.\left(2015-1\right)+2012}{\left(2013+1\right).2015-2016}\)
\(=\frac{2013.2015-2013+2012}{2013.2015+2015-2016}\)
\(=\frac{2013.2015-1}{2013.2015+\left(-1\right)}\)
\(=\frac{2013.2015-1}{2013.2015-1}\)
\(=1\)