Cho A= 1/2^2 +1/3^2 +1/4^2 +...+ 1/2015^2 +1/2016^2
chứng tỏ A <1
K
Khách
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14 tháng 3 2022
Ta có : \(\dfrac{1}{2^2}\)<\(\dfrac{1}{1.2}\); \(\dfrac{1}{3^2}\)<\(\dfrac{1}{2.3}\);.....;\(\dfrac{1}{2016^2}\)<\(\dfrac{1}{2015.2016}\)
⇒ A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{2016^2}\)< \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{2015.2016}\)
⇒ A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{2016^2}\) < 1 - \(\dfrac{1}{2016}\)= \(\dfrac{2015}{2016}\) (ĐCPCM)
9 tháng 8 2016
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2016^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(A< 1-\frac{1}{2016}\)
\(A< \frac{2015}{2016}\left(đpcm\right)\)
9 tháng 8 2016
\(A=\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{2016.2016}< \frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}\)
\(=\frac{2015}{2016}\)
\(\Rightarrow A< \frac{2015}{2016}\)
DN
1
20 tháng 11 2018
\(A=3+3^2+3^3+3^4+...+3^{2015}+3^{2016}\\\)
\(A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2015}\left(1+3\right)\)
\(A=\left(1+3\right).\left(3+3^3+...+3^{2015}\right)\)
\(A=4.\left(3+3^3+...+3^{2015}\right)\)
Suy ra : \(A⋮4\)
DY
0
NV
1
22 tháng 3 2021
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
HD
4
VM
21 tháng 4 2016
1/2.3 +1/3.4+...+1/2016.2017 < 1/2^2+1/3^2+...+1/2016^2
1/2 -1/3+1/3 -1/4+...+1/2016-1/2017 < 1/2^2+1/3^2+...+1/2016^2
1/2-1/2017 < 1/2^2+1/3^2+...+1/2016^2
=> 2015/4034 < 1/2^2+1/3^2+...+1/2016^2
tương tự
1/2^2+1/3^2+...+1/2016^2 < 1/1.2 +1/2.3+...+1/2015.2016
1/2^2+1/3^2+...+1/2016^2 < 1- 1/2+1/2 -1/3+...+1/2015- 1/2016
1/2^2+1/3^2+...+1/2016^2 < 1-1/2016
1/2^2+1/3^2+...+1/2016^2 < 2015/2016
tích nha
ạ
Ta có: \(\frac{1}{n^2}<\frac{1}{n\times\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Từ điều trên, ta có: \(A<\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)
\(A<\frac{1}{2}-\frac{1}{2017}\)
\(A<\frac{2015}{4034}<1\)
0<A<1 nên A không phải là số tự nhiên.
(+)Hiển nhiên A>0 vì các số hạng của A đều > 0 (1)
(+)Tổng quát: \(\frac{1}{n^2}<\frac{1}{\left(n-1\right).n}\)
Ta có:\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2015.2016}\)
\(\Rightarrow A<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}=1-\frac{1}{2016}<1\) (2)
Từ (1);(2)
=>0<A<1
=>A ko là số tự nhiên