Chúng minh:
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{4010^2}< \frac{1}{2}\)
Chứng minh S=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+....................+\frac{1}{4008^2}+\frac{1}{4010^2}< \frac{1}{2}\)
Giúp mình nhé ai nhanh nhất mình tick
chứng minh rằng \(\frac{1}{2^2}\)+ \(\frac{1}{4^2}\)+ \(\frac{1}{6^2}\)+.....+\(\frac{1}{4010^2}\)< \(\frac{1}{2}\)
Ta có:
B=1-1/2²-1/3²-...-1/2004²
=1-(1/2²+1/3²+...+1/2004²)
=1-[1/(2.2)+1/(3.3)+...+1/(2004.2004)]
Ta thấy:
1/(2.2)>1/(2.3)
1/(3.3)>1/(3.4)
...
1/(2004.2004)>1/(2004.2005)
Cộng từng vế của các bất đẳng thức trên ta được:
1/(2.2)+1/(3.3)+...+1/(2004.2004) > 1/(2.3)+1/(3.4)+...+1/(2004.2005) = 1/(3.2)+1/(4.3)+...+1/(2005.2004)
= (3-2)/(3.2)+(4-3)/(4.3)+...+(2005-2004)/(2005.2004)
=3/(3.2)-2/(3.2)+4/(4.3)-3/(4.3)+...+2005/(2005.2004)-2004/(2005.2004)
=1/2-1/3+1/3-1/4+...+1/2004-1/2005
=1/2-1/2005
=2003/4010
=> B>1-2003/4010=2007/4010>2007/4022028=1/2004
Hay B>1/2004
tích nha
cảm ơn bạn nhưng bạn đọc sai đề rồi kìa
bài 1: tính A:=\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{2}{3}-\frac{1}{2}\)
Bài 2: Cho B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{49}-\frac{1}{50}\)
Chứng minh rằng: \(\frac{7}{12}< A< \frac{5}{6}\)
Chứng minh:
a, M= \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\)
b, \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
Chứng minh \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)\(\frac{1}{2}\)
b,\(\frac{1}{^{2^2}}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
c,\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)
d,\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
a, \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)
Ta có: \(\frac{1}{13}< \frac{1}{12};\frac{1}{14}< \frac{1}{12};\frac{1}{15}< \frac{1}{12}\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{61}< \frac{1}{60};\frac{1}{62}< \frac{1}{60};\frac{1}{63}< \frac{1}{60}\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{3}{60}=\frac{1}{20}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
Vậy...
b, Đặt A là tên của tổng trên
Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Đặt B là biêu thức trong ngoặc
Ta có: \(1=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow B< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow B< 2-\frac{1}{50}< 2\)
Thay B vào A ta được:
\(A< \frac{1}{2^2}.2=\frac{1}{2}\)
c, Đặt C = \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)
\(C=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)
\(C=\left(1+1+1+....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)
\(C=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
Đặt D là biểu thức trong ngoặc
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow D< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow-D>-1\)
=>\(C=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>49-1=48\)
Vậy C > 48
Cho \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}\)
CHúng minh tổng trên ko nguyên
Ta gọi tổng trên là B.
Ta luôn có : \(\frac{1}{2^2}>0\)
\(\Rightarrow B>0\left(1\right)\)
Ta có :
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{44.45}\)
\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{44}-\frac{1}{45}\)
\(B< 1-\frac{1}{45}\)
\(B< \frac{44}{45}\)
\(\Rightarrow B< 1\left(2\right)\)
Từ (1) và (2 )
\(\Rightarrow0< B< 1\)
=> Tổng B không nguyên
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{44.45}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}< \frac{44}{45}< 1\)
\(\Leftrightarrow0< \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)
=> Tổng trên không nguyên
chúng minh rằng
1) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}<1\)
2) \(\frac{49}{100}<\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}<1\)
1) Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}< 1\)
1) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 1\)
2)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(...\)
\(\frac{1}{99^2}< \frac{1}{98.99}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{99^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(1-\frac{1}{99}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}< \frac{99}{100}< 1\)
\(\frac{98}{99}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2.3}< \frac{1}{2^2}\)
\(\frac{1}{3.4}< \frac{1}{3^2}\)
\(...\)
\(\frac{1}{99.100}< \frac{1}{99^2}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2}-\frac{1}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{49}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}< 1\)
Chứng minh B = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}< 1\)
Ta có
B = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\) \(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)= \(1-\frac{1}{8}< 1\)
đúng không đấy mình thấy cứ sai sai thế nào ý
mũ hai nhưng \(8^2\) mới lập lại 1 lần
Thank nhưng mình cần kiểm tra lại
chứng minh rằng : s= \(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-......+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+....+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}< 0,2\)
4S=\(\dfrac{4}{2^2}-\dfrac{4}{2^4}+\dfrac{4}{2^6}-...+\dfrac{4}{2^{4n-2}}-\dfrac{4}{2^{4n}}+...+\dfrac{4}{2^{2002}}-\dfrac{4}{2^{2004}}\)
4S=1-\(\dfrac{1}{2^2}+\dfrac{1}{2^4}-,...-\dfrac{1}{2^{2002}}\)
4S+S=1-\(\dfrac{1}{2^{2004}}\)
5S=\(\dfrac{2^{2004}-1}{2^{2004}}\)<1
\(\Rightarrow\)5S<1 hay S<\(\dfrac{1}{5}\)=0,2(đpcm)