Chứng tỏ 1/2^2 + 1/3^2 + 1/4^2 + v.v + 1/1009^2 < 3/4
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
LM
1
S
14 tháng 4 2017
Ta có: \(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)
................
\(\frac{1}{1009^2}< \frac{1}{1008.1009}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{1009^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1008.1009}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1008}-\frac{1}{1009}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{1009}=\frac{3}{4}-\frac{1}{1009}< \frac{3}{4}\)(đpcm)
VT
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
5 tháng 6 2023
A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\)+ \(\dfrac{1}{1+2+3+4}\)+...+\(\dfrac{1}{1+2+3+...+1009}\)
Ta có công thức:
S = 1 + 2 + ...+ n = (n+1)\(\times\)n:2
Áp dụng công thức trên vào A ta có
A = \(\dfrac{1}{\left(2+1\right)\times2:2}\)+\(\dfrac{1}{\left(1+3\right)\times3:2}\)+...+\(\dfrac{1}{\left(1009+1\right)\times1019:2}\)
A = \(\dfrac{1}{2\times3:2}\)+\(\dfrac{1}{3\times4:2}\)+...+\(\dfrac{1}{1009\times1010:2}\)
A = \(\dfrac{2}{2\times3}\)+ \(\dfrac{2}{3\times4}\)+...+\(\dfrac{2}{1009\times1010}\)
A = 2 \(\times\)( \(\dfrac{1}{2\times3}\)+ \(\dfrac{1}{3\times4}\)+...+ \(\dfrac{1}{1009\times1010}\))
A = 2 \(\times\) ( \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+ \(\dfrac{1}{1009}-\dfrac{1}{1010}\))
A = 2 \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{1010}\))
A = 1 - \(\dfrac{2}{1010}\)
A = \(\dfrac{1008}{1010}\)
A = \(\dfrac{504}{505}\)
T
26 tháng 7 2018
Ta có : \(\overline{abcabc}:\overline{abc}=1001\)
\(\Rightarrow\) \(\overline{abcabc}=\overline{abc}\times1001=\overline{abc}\times7\times11\times13\)
Vậy \(\overline{abcabc}\)là bội của 7; 11; 13
____________________________
Tìm x :
a/ \(x+17=-33\)
\(x=-33-17\)
\(x=-50\)
b/ \(2-\left(x-5\right)=5\times2^3\)
\(2-\left(x-5\right)=40\)
\(x-5=2-40\)
\(x-5=-38\)
\(x=-38+5\)
c/ \(1009.x=\left(-1\right)+2+\left(-3\right)+4+\left(-5\right)+6+....+\left(-2017\right)+2018\)
\(1009.x=\left(-1\right)+\left(-1\right)+\left(-1\right)+....+\left(-1\right)\)
\(1009.x=\left(-1\right).1010\)
\(1009.x=-1009\)
\(\Rightarrow x=-1\)
\(x=-33\)
26 tháng 7 2018
1, ta có : abcabc=1000abc +abc=1001abc chia hết cho 7,11,13
2,a,x+17=-33 b,2-(x-5)=5.2^3 c,1009.x=-1+2+(-3)+4+...+(-2017)+2018
x=-33-17 2-(x-5)=5.8 1009.x=(-1+2)+(-3+4)+...+(-2017+2018)
x=-50 2-(x-5)=40 1009x=1+1+1+...+1+1
x-5=-38 1009.x=1009.1
x=-33 1009.x=1009
x=1
PT
0
S
14 tháng 5 2017
a, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2017^2}< \frac{1}{2016.2017}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}>\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}=1-\frac{1}{2017}< 1\)Vậy...
b, Đặt A = \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
Thay B vào A ta được:
\(A< \frac{1}{4}\left(1+1\right)=\frac{1}{4}.2=\frac{1}{2}\)
Vậy....
S
14 tháng 5 2017
c, Ta có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};....;\frac{1}{9^2}>\frac{1}{9.10}\)
\(\Rightarrow A>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)(1)
Lại có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{9^2}< \frac{1}{8.9}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) suy ra \(\frac{2}{5}< A< \frac{8}{9}\)(đpcm)
d, chắc là đề sai
e, giống câu a
20 tháng 3 2023
1:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
...
\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)
=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{7}{8}< 1\)
20 tháng 2 2019
Ta có:
Xét số a. Ta có a2 > (a - 1)(a + 1)
Thật vậy, (a - 1)(a + 1) = a(a + 1) - (a + 1) = a2 + a - a - 1 = a2 - 1 < a2
Suy ra \(\dfrac{1}{\left(a-1\right)\left(a+1\right)}>\dfrac{1}{a^2}\)
Ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(< \dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{100}-\dfrac{1}{101}\right)\)
\(< \dfrac{3}{4}\)
Ko bt có sai chỗ nào ko....
