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\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}< 2\)
\(\Rightarrow A< 2\)
Ta có : \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)\(=1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\)
Vì \(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};..;\frac{1}{50.50}< \frac{1}{49.50}\)nên :
\(\Rightarrow\) \(1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\)\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
Ta có : \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=1+\left(1-\frac{1}{50}\right)\)\(=1+\frac{49}{50}\)
Vì \(\frac{49}{50}< 1\)nên \(1+\frac{49}{50}< 2\)\(\Rightarrow\)\(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)
\(\Rightarrow\)\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)
Bạn xem lời giải ở đường link sau nhé:
Câu hỏi của nguyenducminh - Toán lớp 6 - Học toán với OnlineMath
A=\(\frac{1}{1^2}\)\(+\frac{1}{2^2}\)\(+\frac{1}{3^2}\)\(+...+\frac{1}{50^2}\)
A<1\(+\frac{1}{1.2}\)\(+\frac{1}{2.3}\)\(+...\frac{1}{49.50}\)
=1+1-\(-\frac{1}{2}\)\(+\frac{1}{2}\)\(-\frac{1}{3}\)\(+...+\frac{1}{49}\)\(-\frac{1}{50}\)
=\(1+1-\frac{1}{50}\)
=\(2-\frac{1}{50}\)\(< 2\)
\(\Rightarrow A< 2\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..............+\frac{1}{99^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+................+\frac{1}{98.99}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+............+\frac{1}{98}-\frac{1}{99}\)
\(=1-\frac{1}{99}=\frac{98}{99}< 1\)
\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.............+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...............+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
Vậy \(\frac{49}{100}< A< 1\)
a)=>A=\(1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Đặt tổng trong ngoặc là M
=>M=\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)\(=1-\frac{1}{50}< 1\)
Khi đó A=1+M (M<1)
Ta có công thức :1+x<2 nếu x<1
=>A<1
A=\(\frac{1}{1^2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+............+\(\frac{1}{50^2}\)=\(\frac{1}{4}\)+\(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+........+\(\frac{1}{50.50}\)<\(\frac{1}{4}\)+\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+......+\(\frac{1}{49.50}\)
=\(\frac{1}{4}\)+\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+.........+\(\frac{1}{49}\)-\(\frac{1}{50}\)=\(\frac{1}{4}\)+\(\frac{1}{1}\)-\(\frac{1}{50}\)=\(\frac{123}{100}\)<2
Vay A<2
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}.\)
\(A=1+\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+.......+\frac{1}{50\cdot50}\)
\(< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{49\cdot50}.\)
\(\Rightarrow1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(1+1-\frac{1}{50}< 2\)
=>A<2
ok xong
A = \(\frac{1}{1^2}\) + \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\) + .... + \(\frac{1}{50^2}\)
A = 1 + \(\frac{1}{2.2}\)+ \(\frac{1}{3.3}\)+ \(\frac{1}{4.4}\)+ ...... + \(\frac{1}{50.50}\)< 1 + \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ...... + \(\frac{1}{49.50}\)
A < 1 + ( 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ...... + \(\frac{1}{49}\)- \(\frac{1}{50}\))
A < 1 + ( 1 - \(\frac{1}{50}\))
A < 1 + 1 - \(\frac{1}{50}\)
A < 2 - \(\frac{1}{50}\)
=> A < 2