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\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
Vậy...
1.
Ta có:
Vì b+1-b=1=>\(\dfrac{1}{b}-\dfrac{1}{b+1}=\dfrac{1}{b.\left(b+1\right)}\)<\(\dfrac{1}{b.b}\)(1)
Vì b-(b-1)=1=>\(\dfrac{1}{b-1}-\dfrac{1}{b}=\dfrac{1}{b.\left(b-1\right)}\)>\(\dfrac{1}{b.b}\)(2)
Từ (1) và (2)=>\(\dfrac{1}{b}-\dfrac{1}{b+1}< \dfrac{1}{b.b}< \dfrac{1}{b-1}-\dfrac{1}{b}\)
Câu 2 bạn hỏi bạn Bùi Ngọc Minh nhé PR cho nó
Bài 2:
Ta có:S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{9^2}=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)
S>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\left(1\right)\)
S<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\left(2\right)\)
Từ (1) và (2) suy ra \(\dfrac{2}{5}< S< \dfrac{8}{9}\)
Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
bài 2
| a;đặt biểu thức là S | |
| S < 1/1.2 + 1/2.3 + .......1/(n-1)n | |
| = 1- 1/2 +1 /2 -1/3+........ + 1/n-1 - 1/n | |
|
= 1 -1/n <1 |
|
| vậy S < 1 | |
Đặt A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)
Ta thấy :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\)
\(\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
Nhân xét :
\(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};\)
\(...;\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}+...+\)
\(\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{99}{100}\)
Vì \(A< \dfrac{99}{100}< 1\)
\(\Rightarrow A< 1\)
Bài 1)
Đặt \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)
Ta thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4};....;\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)\(\Rightarrow\) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
\(\Rightarrow\) A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow\) A < \(1-\dfrac{1}{100}\) < 1 \(\Rightarrow\) A < 1
Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)< 1
Ta thấy A > 0 (1)
Vì \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2016^2}< \dfrac{1}{2015.2016}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2015.2016}\)
\(\Rightarrow A>1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2015}-\dfrac{1}{2016}=1-\dfrac{1}{2016}=\dfrac{2015}{2016}< 1\)(2)
Từ (1)(2) => 0 < A < 1
Vậy A không phải là số tự nhiên
Giải:
Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}>0_{\left(1\right)}.\) (do A là phân số dương).
Ta lại có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}.\)
\(=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2016.2016}.\)
\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}.\)
\(< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}.\)
\(< 1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{2015}-\dfrac{1}{2015}\right)-\dfrac{1}{2016}.\)\(< 1+0+0+0+...+0-\dfrac{1}{2016}.\)
\(< 1-\dfrac{1}{2016}.\)
\(< \dfrac{2015}{2016}.\)
\(\Rightarrow A< 1_{\left(2\right)}.\) (do \(\dfrac{2015}{2016}< 1\)).
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\) \(\Rightarrow0< A< 1.\)
\(\Rightarrow A\) không phải là số tự nhiên.
Vậy ta thu được \(đpcm.\)
~ Học tốt!!! ~
1)Ta thấy: \(\dfrac{1}{n^2}=\dfrac{1}{n.n}< \dfrac{1}{\left(n-1\right)n}\)
=>A=\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}...+\dfrac{1}{50^2}< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49.50}\)
A<\(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)
Vậy A<2
2)Ta có:2S=6+3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\)
2S-S=(6+3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\))-(3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\))
=>S=6-\(\dfrac{3}{2^9}=\dfrac{6.2^9-3}{2^9}\)
Vậy S=\(\dfrac{6.2^9-3}{2^9}\)
A = \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}\)
A < \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
A < \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A < \(\dfrac{1}{2}-\dfrac{1}{100}\)
⇒ A < \(\dfrac{1}{2}\)