A= \(\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\) CMR A < 2
CMR :a) \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+.....+\frac{1}{100^2}< \frac{1}{2}\)
b) \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+....+\frac{2499}{2500}>48\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
TA có :\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\left(đpcm\right)\)
A= \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+....+\frac{2499}{2500}\)
A=\(1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+....+1-\frac{1}{2500}\)
A=\(\left(1+1+1+.....+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
A=\(49-\)\(\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
do \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)>0 nên 49<0
bài trên iu cầu CMR A < 49 thì mk lm đúng chưa ạ. Đây là đề thi quận mk đó ạ
(: ko bít. tui giỏi tiếng anh nhưng ngu toán lắm
Bài 1:CMR:\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{15}<2\)
Bài 2: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right)}=\frac{99}{101}\)
Bài 3:\(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2449}{2500}\)
Bài 4:CMR:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Chứng minh rằng :
a) \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
b) \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)
a) Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
. . .
\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2^2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\right)\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(1+1-\frac{1}{50}\right)\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\cdot\frac{99}{50}=\frac{99}{200}< \frac{100}{200}=\frac{1}{2}\left(đpcm\right)\)
b) Ta có :
\(B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)
\(\Rightarrow1-\frac{1}{4}+1-\frac{1}{9}+...+1-\frac{1}{2500}>48\)
\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< 49\)
Lại có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
. . .
\(\frac{1}{50^2}< \frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{50^2}< \frac{49}{50}< 1\)
\(\Rightarrow-\left(\frac{1}{2^2}+...=\frac{1}{50^2}\right)>1\)
\(\Rightarrow49-\left(\frac{1}{2^2}+...+\frac{1}{50^2}\right)>49-1=48\)
hay \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}>48\left(đpcm\right)\)
CHỨNG MINH:
\(a,A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}< 1\)
\(b,B=\frac{1}{3^2}+\frac{1}{9^2}+...+\frac{1}{409^2}< \frac{1}{12}\)
\(c,C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}+...+\frac{2499}{2500}\) KHÔNG PHẢI LÀ SỐ TỰ NHIÊN
THANKS :33
A=\(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{100}\right)\)
B=\(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)+\left(1+\frac{1}{15}\right)......\left(1+\frac{1}{100}\right)\)
D=\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2499}{2500}\)
Tính giá trị các biểu thức:
a)\(A=\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{195\cdot199}\)
b)\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}\)
c)\(C=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
a, \(A=\frac{12}{3.7}+\frac{12}{7.11}+...+\frac{12}{195.199}\)
\(=3.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{195.199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{195}-\frac{1}{199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{199}\right)\)
\(=3.\left(\frac{199}{597}-\frac{3}{597}\right)\)
\(=3.\frac{196}{597}\)
\(=\frac{196}{199}\)
Bài 5 : Chững minh rẳng :
a) S= \(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) CMR :1< S <2
b) \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
c) \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)
d) \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)
Bài :So sánh phân số sau:
a)\(\frac{1985.1987-1}{1980+1985.1986}và1\)
b) A= \(\frac{13^{15}+1}{13^{16}+1}\)và B = \(\frac{13^{16}+1}{13^{17}+1}\)
c)\(\frac{18}{53}và\frac{26}{79}\)
d)\(\frac{5}{8}và\frac{14}{17}\)
e)\(\frac{1}{5^{199}}và\frac{1}{3^{300}}\)
g)\(\frac{1}{3^{17}}và\frac{1}{5^{10}}\)
h) \(\frac{18}{109}và\frac{5}{30}\)
CMR: A > \(\frac{65}{132}\)
A = \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}\)
\(\Rightarrow A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
\(A>1+0+0+0+...+0-\frac{1}{10}\)
\(A>1-\frac{1}{10}=\frac{9}{10}\)
\(\Rightarrow A>\frac{5}{10}=\frac{1}{2}\)
mà \(\frac{1}{2}=\frac{66}{132}>\frac{65}{132}\)
\(\Rightarrow A>\frac{65}{132}\)
Vậy \(A>\frac{65}{132}\)
Ta có : \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(A=\frac{1}{4}+\frac{1}{3^2}+...+\frac{1}{10^2}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}\)
\(\Rightarrow A>\frac{65}{132}\)
Vậy \(A>\frac{65}{132}\) \(\left(đpcm\right)\)