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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
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:
\(A< \frac{1}{2^2}.2=\frac{1}{2}\)
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)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(...\)
\(\frac{1}{99^2}< \frac{1}{98.99}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{99^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(1-\frac{1}{99}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}< \frac{99}{100}< 1\)
\(\frac{98}{99}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2.3}< \frac{1}{2^2}\)
\(\frac{1}{3.4}< \frac{1}{3^2}\)
\(...\)
\(\frac{1}{99.100}< \frac{1}{99^2}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{1}{2}-\frac{1}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\frac{49}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}< 1\)
A=\(1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Đặt B=\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+..+\)\(\frac{1}{99.100}=\)\(1-\frac{1}{100}< 1\)
Mà A=1+B=>A=1+B<1+1=2
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
vậy \(A=\frac{99}{100}< 2\left(đpcm\right)\)
B)
ta có : \(1=1\)
\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)
\(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{7}< \frac{1}{4}+...+\frac{1}{4}=\frac{4}{4}=1\)
\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}< \frac{1}{8}+...+\frac{1}{8}=\frac{8}{8}=1\)
\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{63}< 1\)
tất cả công lại \(\Rightarrow B< 6\)
b, Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
..................
\(\frac{1}{100^2}< \frac{1}{99.100}\)
Nên C < \(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\)
<=> C < \(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)
<=> C < \(1+1-\frac{1}{100}\)
<=> C < \(2-\frac{1}{100}=\frac{199}{100}\)
\(B=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+...+\left(\frac{1}{2^5}+...+\frac{1}{2^6-1}\right)\)
\(B< 1+\frac{1}{2}.2+\frac{1}{4}.4+...+\frac{1}{2^5}.32\)
\(B< 1+1+1+...+1\)( 6 số 1)
B<1.6=6
\(C=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
\(C< 1+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.10}\right)=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)\(=1+\left(1-\frac{1}{100}\right)< 1+1=2\)
Vậy C<2
Đặt A là biểu thức của đề bài
Ta có \(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2005^2}\right)\)
\(A< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}\right)\)
\(A< \frac{1}{2^2}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2004}-\frac{1}{2005}\right)\)
\(A< \frac{1}{2^2}\left(1+1-\frac{1}{2005}\right)< \frac{1}{2^2}\left(1+1\right)=\frac{1}{2^2}.2=\frac{1}{2}\)
Vậy A<1/2
Cảm ơn bạn Tuấn Minh