Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{7\cdot8}\)
Dễ thấy: \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)\(< A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{7\cdot8}\left(1\right)\)
Ta có:\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{7\cdot8}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(=1-\dfrac{1}{8}< 1\left(2\right)\)
Từ \((1);(2)\) ta có: \(B< A< 1\Rightarrow B< 1\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}\)
vì \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
...............
\(\frac{1}{8^2}< \frac{1}{7\cdot8}\)
nên \(B< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow B< 1-\frac{1}{8}\)
\(\Rightarrow B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\)
ta có :
\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)
\(..........................\)
\(\frac{1}{8^2}=\frac{1}{8.8}< \frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{7.8}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow B< 1-\frac{1}{8}\)
\(\Rightarrow B< \frac{7}{8}\) ( 1 )
mà \(\frac{7}{8}< 1\) ( 2 )
từ ( 1 ) và ( 2 ) \(\Rightarrow B< 1\)
vậy ......................
có : 1/2^2 < 1/1*2; 1/3^2 < 1/2*3; 1/4^2 < 1/3*4;....; 1/8^2 < 1/7*8
=> B < 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/7*8
=> B < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/7 - 1/8
=> B < 1 - 1/8
=> B < 7/8 < 1
=> B < 1
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}\)
\(B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{8.8}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow B< 1-\frac{1}{8}\)
\(\Rightarrow B< \frac{7}{8}\)
Mà : \(\frac{7}{8}< 1\)
\(\Rightarrow B< 1\)
Vậy : B < 1
Ta có: \(\frac{1}{1^2}=1\)
\(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
...
\(\frac{1}{50^2}<\frac{1}{49.50}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
=> A < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
=> A < 1 - 1/50 = 49/50
Mà 49/50 < 50/50 = 1 < 2
=> A < 2 (Đpcm).
Ta có:
1/2^2<1/1.2
1/3^2<1/2.3
...
1/1007^2<1/1006.1007
=>1/2^2+1/3^2+...+1/1007^2<1/1.2+1/2.3+...+1/1006.1007
=>1/2^2+1/3^2+...+1/1007^2<1-1/2+1/2-1/3+...+1/1006-1/1007
=>1/2^2+1/3^2+...+1/1007^2<1-1/1007<1
=>1/2^2+1/3^2+...+1/1007^2<1
=> (1/42 + 1/62 + 1/82 + ... + 1/20142).4<1/4.4
1/42 + 1/62 + 1/82 + ... + 1/20142 <1/4
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{25^2}< \dfrac{1}{24\cdot25}=\dfrac{1}{24}-\dfrac{1}{25}\)
Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{25^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{25^2}< 1-\dfrac{1}{25}\)
=>\(1+\dfrac{1}{2^2}+...+\dfrac{1}{25^2}< 2-\dfrac{1}{25}\)
=>\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{25^2}\right)< \dfrac{1}{4}\left(2-\dfrac{1}{25}\right)=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)