A=\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2014\cdot2015\cdot2016}=\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2014\cdot2015}-\dfrac{1}{2015\cdot2016}\right)=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2015}\cdot\dfrac{1}{2016}\right)=\dfrac{1}{4}-\dfrac{1}{2\cdot2015\cdot2016}< \dfrac{1}{4}\)

Vậy A<\(\dfrac{1}{4}\)

---bé hơn hoặc bằng tức là chỉ cần xảy ra 1 khả năng cũng thõa mãn nhé---

mình

bài này tương tự bài trên

Đặt \(A=1.2.3+2.3.4+3.4.5+...+2015.2016.2017\)

=>\(4A=1.2.3.4+2.3.4.4+3.4.5.4+...+2015.2016.2017.4\)

=>\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)\)

\(+...+2015.2016.2017.\left(2018-2014\right)\)

=>\(4A=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5\)

\(+...+2015.2016.2017.2018-2014.2015.2016.2017\)

=>\(4A=2015.2016.2017.2018\Rightarrow A=\frac{2015.2016.2017.2018}{4}\)

Ta có:

\(9911=11\cdot17\cdot53\)

Để \(A=1.3.5...2015+2.4.6....2016⋮9911\)thì:\(\hept{\begin{cases}1.3.5...2015⋮9911\\2.4.6...2016⋮9911\end{cases}}\)

Mà: \(1.3.5...2015=1.3.5...11.13.15.17...53...2015⋮11.17.53=9911\)

và \(2\cdot4\cdot...\cdot2016=2\cdot4\cdot...\cdot22\cdot...\cdot34\cdot...\cdot106\cdot...\cdot2016⋮11\cdot17\cdot54=9911\)

=> đpcm

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)

\(A=\frac{1}{4.6}+\frac{1}{10.12}+\frac{1}{18.20}+...+\frac{1}{810.812}\)

.......

~ Chúc học tốt ~ 

Ai ngang qua xin để lại 1 L - I - K - E

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{27.28.29.30}\)

\(3A=3.\left(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{1}{27.28.29.30}\right)\)

\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+..........+\frac{3}{27.28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)

\(3A=\frac{1}{6}-\frac{1}{24360}\)

\(3A=\frac{1353}{8120}\)

\(A=\frac{1353}{8120}:3\)

\(A=\frac{451}{8120}\)

Ta có:3A=\(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+.............+\frac{3}{27.28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)

\(3A=\frac{1353}{8120}\Rightarrow A=\frac{451}{8120}\)

Vậy \(A=\frac{8056}{2015}\)

Bạn tham khảo: Câu hỏi của chipchip - Toán lớp 6 - Học toán với OnlineMath

\(A=\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2014\cdot2015}\)

\(\frac{1}{4}A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\)

\(\frac{1}{4}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2014}-\frac{1}{2015}\)

\(\frac{1}{4}A=1-\frac{1}{2015}\)

\(\frac{1}{4}A=\frac{2014}{2015}\)

\(A=\frac{2014}{2015}:\frac{1}{4}=\frac{2014}{2015}\cdot4=\frac{8056}{2015}\)

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+....+\frac{1}{47.48.49.50}\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{47.48.49}-\frac{1}{48.49.50}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{48.49.50}\right)\)

\(=\frac{1}{3}.\frac{6533}{39200}=\frac{6533}{117600}\)