Cho A =1/1^2+1/2^2+1/3^2+……+1/50^2
Cm A<2
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\(Cm:\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)< 2
Ta có: \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{49}{50}< 1< 2\)
=> A < 2
tk nha mn
Ta có: \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\) \(=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\) \(=1+\left(\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\right)< 1+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)\)
\(\Rightarrow A< 1+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(\Rightarrow A< 1+\left(\frac{1}{2}-\frac{1}{51}\right)=1+\frac{49}{102}< 1+1=2\) (Đpcm)
\(\Rightarrow A<1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{49.50}\)
\(\Rightarrow A<1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow A<1+\left(1-\frac{1}{50}\right)\)
\(\Rightarrow A<1+\frac{49}{50}\)
\(\Rightarrow A<\frac{99}{50}\)
Vì \(\frac{99}{50}<2=\frac{100}{50}\Rightarrow A<2\) ĐPCM
Ta có:
\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};......;\frac{1}{50^2}<\frac{1}{49.50}\)
Do đó \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}<1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(\Rightarrow A<1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}<2\)
=>A<2(đpcm)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(A=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)=1+B\)( Gọi biểu thức trong ngoặc là B)
Ta xét B
B=\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
B<\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
B<\(1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+...+\frac{1}{49}-\frac{1}{50}\)
B<\(1-\frac{1}{50}<1\)
Vậy B<1
=>A=1+B < 1+1=2
Vậy A<2
Có A = 1/12 + 1/22+ 1/32+ ...+ 1/502 => A< 1/12 + 1/1*2 + 1/2*3 + 1/3*4+ ...+ 1/49*50 A< 1+ 1- 1/2+ 1/2- 1/3 + 1/3- 1/4+ ...+ 1/49 - 1/50 A< 1+ 1-1/50 = 1+ 49/50. Mà 1+49/50 < 1+1=2. => A<2 (ĐPCM)