P=3 /1.2² +1/2².3²+...+4033/2016².2017²

P=1/1 -1/2² +1/2² -1/5² +...+1/2016² - 1/2017²

P=1-1/2017² <1

vậy p<1

Ta có nhận xét sau 

Với n là số nguyên dương

\(\frac{2n+1}{n^2.\left(n+1\right)^2}=\frac{n^2+2n+1-n^2}{n^2\left(n+1\right)^2}=\frac{\left(n+1\right)^2}{n^2\left(n+1\right)^2}-\frac{n^2}{n^2\left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)

Áp dụng nhận xét trên ,ta có 

\(P=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{2016^2}-\frac{1}{2017^2}=1-\frac{1}{2017^2}>1\)

\(P=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+.....+\frac{4033}{\left(2016.2017\right)^2}\)

\(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+.....+\frac{2017^2-2016^2}{2016^2.2017^2}\)

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+....+\frac{1}{2016^2}-\frac{1}{2017^2}\)

\(=1-\frac{1}{2017^2}< 11\) (đpcm)

Bài này trong đề thi học kì 2 môn Toán lớp 6 trường Amsterdam năm 2016-2017 này. Mình 10 luôn hehe

Ta có \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2016.2017}\)

\(\Rightarrow A=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)

\(\Rightarrow A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2016}+\frac{1}{2017}\right)\)

\(\Rightarrow A=2\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=2\left(\frac{2016}{2017}\right)\)

\(\Rightarrow A=\frac{4032}{2017}\)

Ta có:\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+....+\frac{2}{2016\cdot2017}\)

\(=\frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+....+\frac{2}{2016}-\frac{2}{2017}\)

\(=\frac{2}{1}-\frac{2}{2017}=2-\frac{2}{2017}=\frac{4034}{2017}-\frac{2}{2017}=\frac{4032}{2017}\)

\(A=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{2016\cdot2017}\)

\(\frac{A}{2}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)

\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(\frac{A}{2}=1-\frac{1}{2017}=\frac{2016}{2017}\)

\(A=\frac{2016}{2017}\cdot2=\frac{4032}{2017}\)

Đối với dạng này ta dùng công thức \(a\cdot\left(a+1\right)=\dfrac{1}{3}\left[a\cdot\left(a+1\right)\cdot\left(a+2\right)-\left(a-1\right)\cdot a\cdot\left(a+1\right)\right]\)

Ta có:

\(1\cdot2=\dfrac{1}{3}\left(1\cdot2\cdot3-0\cdot1\cdot2\right)\)

\(2\cdot3=\dfrac{1}{3}\left(2\cdot3\cdot4-1\cdot2\cdot3\right)\)

$\cdots$

\(2016\cdot2017=\dfrac{1}{3}\left(2016\cdot2017\cdot2018-2015\cdot2016\cdot2017\right)\)

Cộng lại ta có: \(1\cdot 2 +2\cdot 3 +3 \cdot 4 +\cdots +2016\cdot 2017=\dfrac{1}{3} (2016\cdot 2017 \cdot 2018-0\cdot 1 \cdot 2)=\dfrac{1}{3}\cdot 2016\cdot 2017 \cdot 2018 \)

Thay vào $A$ thu được $A=672.$

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

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right):2\)

\(=\left(1-\frac{1}{2017}\right):2\)\(< \)\(\frac{1}{2}\)   (Do 1 - 1/2017 < 1)

A=1.2+2.3+3.4+4.5+5.6+...+2016.2017

=> 3A = 1.2.3+2.3.3+3.4.3+4.5.3+5.6.3+.......+2016.2017.3

=> 3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + 4.5.(6-3) + .......+ 2016.2017.(2018-2015)

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +..........+ 2016.2017.2018 - 2015.2016.2017

=> 3A = 2016.2017.2018

=> A =  2016.2017.2018 : 3

Ta thấy:Các số trong dãy số trên cách nhau 1,1 đơn vị.

Số các số hạng là:

       ( 2016,2017 - 1,2 ) : 1,1 + 1 = 1832,819727 ( số )

Tổng là:

        ( 2016,2017 + 1,2 ) x 1832,819727 : 2 = 1848766,817

                              Đ/S: số trên dài wóa :))