Ta có: 1/22 < 1/1.2
1/32 < 1/2.3
1 /4 2 < 1/3.4
.. .........................
1/502 < 1/49.50
=> A < 1/12 + 1/1.2 + 1/2.3 + 1/3.4+......+1/49.50
=> A < 1 + (1-1/50)
=> A < 1+49/50
=> A < 99/55 <2
=> A < 2
Ta có: 1/22 < 1/1.2
1/32 < 1/2.3
1 /4 2 < 1/3.4
.. .........................
1/502 < 1/49.50
=> A < 1/12 + 1/1.2 + 1/2.3 + 1/3.4+......+1/49.50
=> A < 1 + (1-1/50)
=> A < 1+49/50
=> A < 99/55 <2
=> A < 2
\(A<1+\frac{1}{1.2}+\frac{1}{2.3}+..............+\frac{1}{49.50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.................+\frac{1}{49}-\frac{1}{50}\)
\(=2-\frac{1}{50}<2\)
\(\Leftrightarrow A<2\)
Số số hạng của A là : (50-1):1+1=50 số
Vì :
\(\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{2.3}\)
\(\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3}\)
\(...\)
\(\frac{1}{50^2}=\frac{1}{50.50}<\frac{1}{49.50}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}<1-\frac{1}{50}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}<\frac{49}{50}\)
Vì \(\frac{49}{50}<1\)
Mà 1\(\frac{1}{1^2}=\frac{1}{1.1}=\frac{1}{1}=1\)
\(\Rightarrow A<1+1\)
\(\Leftrightarrow A<2\)
Vậy A < 2
Ta có: 1/22 < 1/1.2
1/32 < 1/2.3
1 /4 2 < 1/3.4
.. .........................
1/502 < 1/49.50
=> A < 1/12 + 1/1.2 + 1/2.3 + 1/3.4+......+1/49.50
=> A < 1 + (1-1/50)
=> A < 1+49/50
=> A < 99/55 <2
=> A < 2
Ta có: 1/22 < 1/1.2
1/32 < 1/2.3
1 /4 2 < 1/3.4
.. .........................
1/502 < 1/49.50
=> A < 1/12 + 1/1.2 + 1/2.3 + 1/3.4+......+1/49.50
=> A < 1 + (1-1/50)
=> A < 1+49/50
=> A < 99/55 <2
=> A < 2
