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mỗi p/số của A đều bé hơn 1/1.2+1/2.3+1/3.4+......+1/49.50
A<1-1/2+1/2-1/3+1/3-1/4+..........+1/49-1/50(tách ra thành hiệu)
A<1-1/50
mà 1/50>0=>1-1/50<1<2
A<1-1/50<1<2
A<2
chúc học tốt
Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2015.2015}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(=1-\frac{1}{2015}=\frac{2014}{2015}< 1\)
=> A < 1 (đpcm)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< 1-\frac{1}{2017}=\frac{2016}{2017}>\frac{1}{2}\)
\(\Rightarrow\)ko thể cm
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}\)
Ta có :\(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}=\frac{1}{2.3}\)
.........
\(\frac{1}{2017^2}=\frac{1}{2016.2017}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{2}-\frac{1}{2017}\)
\(A=\frac{-1}{4}-\frac{1}{2017}=\frac{-2021}{8068}\)
\(\Leftrightarrow A< \frac{1}{2}\) . Vì \(\frac{-2021}{8068}< \frac{1}{2}\)
A=1+[\(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}\)
ta có \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};......;\frac{1}{50^2}<\frac{1}{49.50}\)
=>A<1+\(\left[\frac{1}{1.2}+.........+\frac{1}{49.50}\right]\)
=>A<1+\(\left[\frac{1}{1}-\frac{1}{50}\right]\)
=>A<1+\(\frac{49}{50}\)
=>A<\(\frac{99}{50}\) <2
=>A<2
K MÌNH NHA BÀI NÀY MÌNH GHI MỎI TAY LẮM
A=\(\frac{1}{1^2}+\frac{1}{2^2}+....+\frac{1}{50^2}\)
A<\(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49\cdot50}\)
A<1+\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
A<1+\(\left(1-\frac{1}{50}\right)\)
A<1+\(\frac{49}{50}\)
=>A<2
mình chỉ gợi ý thôi, vì viết cái này mỏi tay lắm thông cảm nha
Ở phần ''a'' bạn hãy đổi ra thành:2=2;4=2;.....sau dó bạn CM \(\frac{1}{2^2}<\frac{1}{1.2}.....\) rồi hãy suy ra nhỏ hơn \(\frac{1}{3}\)
còn phần ''b'' bạn hãy tách ra nha
Ta có A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^100
Suy ra 2A - A = ( 1 + 1/2 + 1/2^2 +...+ 1/2^99) - ( 1/2 + 1/2^2 +...+ 1/2^100 )
Suy ra A = 1 - 1/2^100 < 1
Vậy A < 1 ( ĐPCM)
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)
\(B< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\)
\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
\(B< 1-\frac{1}{8}< 1\left(dpcm\right)\)
Ta có :
\(\frac{1}{3^2}< \frac{1}{2\times3};\frac{1}{4^2}< \frac{1}{3\times4};\frac{1}{5^2}< \frac{1}{4\times5};\frac{1}{6^2}< \frac{1}{5\times6};...;\frac{1}{100^2}< \frac{1}{99\times100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{99\times100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
\(A<\frac{1}{1\cdot2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}<1<2\)