Ta có: k(k + 1)(k + 2) = 1/4. k(k + 1)(k + 2). 4
= 1/4. k(k + 1)(k + 2). [(k + 3) - (k - 1)]
= 1/4. k(k + 1)(k + 2)(k + 3) - 1/4. k(k + 1)(k + 2)(k - 1)
=> 4S = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + k(k + 1)(k + 2)(k + 3) - k(k + 1)(k + 2)(k - 1)
= k(k + 1)(k + 2)(k + 3)
=> 4S + 1 = k(k + 1)(k + 2)(k + 3) + 1
Đây là tổng của 4 số liên tiếp cộng 1 nên luôn là số chính phương.

cam on rat nhieu

\(A=\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{2015.2016.2017}\)

\(A=\frac{3}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+\frac{3}{2}.\left(\frac{1}{3.4}-\frac{1}{4.5}\right)+...+\frac{3}{2}.\left(\frac{1}{2015.2016}-\frac{1}{2016.2017}\right)\)

\(A=\frac{3}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\right)\)

\(A=\frac{3}{2}.\left(\frac{1}{1.2}-\frac{1}{2016.2017}\right)\)

\(A=\frac{3}{4}-\frac{3}{2.2016.2017}< 1\)

Ta có: \(S=1\cdot2\cdot3+2\cdot3\cdot4+\cdots+9\cdot10\cdot11\)

\(=2\left(2-1\right)\left(2+1\right)+3\left(3-1\right)\left(3+1\right)+\cdots+10\left(10-1\right)\left(10+1\right)\)

\(=2\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+10\left(10^2-1\right)\)

\(=\left(2^3+3^3+\cdots+10^3\right)-\left(2+3+\cdots+10\right)\)

\(=\left(1^3+2^3+\cdots+10^3\right)-\left(1+2+3+\cdots+10\right)\)

\(=\left(1+2+3+\cdots+10\right)^2-\left(1+2+3+\cdots+10\right)\)

\(=\left(10\cdot\frac{11}{2}\right)^2-10\cdot\frac{11}{2}=55^2-55\)

=>\(4S=4\cdot55^2-4\cdot55\)

=>4S+1=\(4\cdot55^2-4\cdot55+1=\left(2\cdot55\right)^2-2\cdot2\cdot55\cdot1+1^2=\left(2\cdot55-1\right)^2\)
=>4S+1 là số chính phương

4N = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 2015.2016.2017.(2018-2014)

4N = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 2015.2016.2017.2018 - 2014.2015.2016.2017

4N = (1.2.3.4 + 2.3.4.5 + 3.4.5.6 + ... + 2015.2016.2017.2018) - (0.1.2.3 + 1.2.3.4 + 2.3.4.5 + ... + 2014.2015.2016.2017)

4N = 2015.2016.2017.2018 - 0.1.2.3

4N = 2015.2016.2017.2018

N = 2015.2016.504.2018 (kq hơi to nên bn tự tính nhé)

Ta có \(k\left(k+1\right)\left(k+2\right)=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\cdot4\)

\(=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left[\left(k+3\right)-\left(k-1\right)\right]\\ =\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left(k+3\right)-\dfrac{1}{4}\left(k-1\right)k\left(k+1\right)\left(k+2\right)\)

Từ đó ta được \(S=\dfrac{1}{4}\cdot1\cdot2\cdot3\cdot4-\dfrac{1}{4}\cdot0\cdot1\cdot2\cdot3+...+\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12-\dfrac{1}{4}\cdot8\cdot9\cdot10\cdot11\\ \Leftrightarrow S=\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12\\ \Leftrightarrow4S+1=9\cdot10\cdot11\cdot12+1=11881=109^2\left(đpcm\right)\)

\(S=1.2.3+2.3.4+3.4.5+...+9.10.11\)

\(4S=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+9.10.11.\left(12-8\right)\)

\(=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+9.10.11.12-8.9.10.11\)

\(=9.10.11.12\)

\(4S+1=9.10.11.12+1=\left(9.12\right).\left(10.11\right)+1=108.110+1\)

\(=\left(109-1\right)\left(109+1\right)+1=109^2-1+1=109^2\)

Ta có đpcm. 

Ta có \(k\left(k+1\right)\left(k+2\right)=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\cdot4\)

\(=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left[\left(k+3\right)-\left(k-1\right)\right]\\ =\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left(k+3\right)-\dfrac{1}{4}\left(k-1\right)k\left(k+1\right)\left(k+2\right)\)

Từ đó ta được \(S=\dfrac{1}{4}\cdot1\cdot2\cdot3\cdot4-\dfrac{1}{4}\cdot0\cdot1\cdot2\cdot3+...+\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12-\dfrac{1}{4}\cdot8\cdot9\cdot10\cdot11\\ \Leftrightarrow S=\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12\\ \Leftrightarrow4S+1=9\cdot10\cdot11\cdot12+1=11881=109^2\left(đpcm\right)\)