CMR: 3/1^2. 2^2 +5/2^2. 3^2+..........+19/9^2. 10^2 < 1
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ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều
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\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2+3^2}+...+\dfrac{19}{9^2-10^2}\)
\(=\) \(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=\) \(1-\dfrac{1}{10^2}< 1\) ( đpcm )
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\(\frac{3}{1^2+2^2}+\frac{5}{2^2+3^2}+...+\frac{19}{9^2+10^2}=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{19}{81.100}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}=1-\frac{1}{100}< 1.\)
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Ta có :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=1-\frac{1}{10^2}< 1\)
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\(\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+...+\frac{19}{9^2\cdot10^2}\)\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{9^2}-\frac{1}{10^2}=1-\frac{1}{10^2}=\frac{99}{100}\)<1