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

\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2016.2016}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(=1-\frac{1}{2016}\)

\(=\frac{2015}{2016}< 1\)

\(\Rightarrow A< 1\)

\(\text{Vậy }A< 1\left(\text{đpcm}\right)\)

                                                                     Bài giải

 Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\) ;  \(\frac{1}{3^2}< \frac{1}{2\cdot3}\) ; \(\frac{1}{4^2}< \frac{1}{3\cdot4}\)  ; ... ; \(\frac{1}{2016^2}< \frac{1}{2015\cdot2016}\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}=1-\frac{1}{2016}=\frac{2015}{2016}< 1\)

                                   \(\Rightarrow\text{ }A< 1\)

                                                                     Bài giải

 Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\) ;  \(\frac{1}{3^2}< \frac{1}{2\cdot3}\) ; \(\frac{1}{4^2}< \frac{1}{3\cdot4}\)  ; ... ; \(\frac{1}{2016^2}< \frac{1}{2015\cdot2016}\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}=1-\frac{1}{2016}=\frac{2015}{2016}< 1\)

                                   \(\Rightarrow\text{ }A< 1\text{ }\left(\text{ ĐPCM}\right)\)

a)\(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}<1\)

\(\Rightarrow2M=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}<1\)

\(2M-M=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}\right)<1\)

\(\Rightarrow M=1-\frac{1}{2016^2}\)<1

=>(DPCM)

CÂU b và c làm tương tự

chtt 

nhé bn

2.

Ta có : \(A=\frac{n+5}{n+2}=\frac{n+2+3}{n+2}=1+\frac{3}{n+2}\)

để A là số nguyên thì \(\frac{3}{n+2}\)là số nguyên

\(\Rightarrow3⋮n+2\)

\(\Rightarrow\)n + 2 \(\in\)Ư ( 3 ) = { 1 ; -1 ; 3 ; -3 }

Lập bảng ta có :

Vậy n \(\in\){ -1 ; -3 ; 1 ; -5 }

3. 

\(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\)

\(=\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+\left(1+\frac{1}{27}\right)+...+\left(1+\frac{1}{3^{98}}\right)\)

\(=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^{98}}\right)\)

\(=97+\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

gọi \(B=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)( 1 )

\(3B=1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)( 2 )

Lấy ( 2 ) trừ ( 1 ) ta được :

\(2B=1-\frac{1}{3^{98}}< 1\)

\(\Rightarrow B=\frac{1-\frac{1}{3^{98}}}{2}< \frac{1}{2}< 1\)

\(\Rightarrow97+\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)< 100\)

4.

đặt \(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)

\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\)

\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\)

\(5A=1-\frac{1}{31}< 1\)

\(\Rightarrow A=\frac{1-\frac{1}{31}}{5}< \frac{1}{5}< 1\)

Ta có : \(2A=2.\left(1+2+2^2+2^3+...+2^{2015}+2^{2016}\right)\)

            \(2A=2+2^2+2^3+2^4+...+2^{2016}+2^{2017}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2016}+2^{2017}\right)-\left(1+2+2^2+2^3+...+2^{2015}+2^{2016}\right)\)

\(A=2+2^3+2^4+2^5+...+2^{2016}+2^{2017}-1-2-2^2-2^3-...-2^{2015}-2^{2016}\)

\(A=2^{2017}-1\)

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

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}+\frac{1}{2^{2017}}\right)\)

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

\(=>đpcm\)

Ủng hộ mk nha ^_-