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\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
\(a)\) Ta có :
\(\frac{1}{100}A=\frac{100^{2009}+1}{100^{2009}+100}=\frac{100^{2009}+100}{100^{2009}+100}-\frac{99}{100^{2009}+100}=1-\frac{99}{100^{2009}+100}\)
\(\frac{1}{100}B=\frac{100^{2010}+1}{100^{2010}+100}=\frac{100^{2010}+100}{100^{2010}+100}-\frac{99}{100^{2010}+100}=1-\frac{99}{100^{2010}+100}\)
Vì \(\frac{99}{100^{2009}+100}>\frac{99}{100^{2010}+100}\) nên \(1-\frac{99}{100^{2009}+100}< 1-\frac{99}{100^{2010}+100}\)
Do đó :
\(\frac{1}{100}A< \frac{1}{100}B\)\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
\(A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A>\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{100.101}\)
\(A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{101}\)
\(A>\frac{1}{5}-\frac{1}{101}=\frac{100}{505}>\frac{100}{600}=\frac{1}{6}\)
Tương tự
\(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)
\(A< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
:\(\frac{1}{6}\)<\(\frac{1}{5^2}\)+\(\frac{1}{6^2}\)+\(\frac{1}{7^2}\)+.....+\(\frac{1}{100^2}\)<\(\frac{1}{4}\)
=\(\frac{1}{5^2}\)+\(\frac{1}{6^2}\)+\(\frac{1}{7^2}\)+.....+\(\frac{1}{100^2}\)<\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+.....+\(\frac{1}{99.100}\)
=\(\frac{1}{4}\)-\(\frac{1}{100}\)=\(\frac{24}{100}\)<\(\frac{25}{100}\)=\(\frac{1}{4}\)>\(\frac{20}{100}\)=\(\frac{1}{5}\)>\(\frac{1}{6}\)
Vậy \(\frac{1}{6}\)<\(\frac{1}{5}\)
:$\frac{1}{6}$16 <$\frac{1}{5^2}$152 +$\frac{1}{6^2}$162 +$\frac{1}{7^2}$172 +.....+$\frac{1}{100^2}$11002 <$\frac{1}{4}$14
=>$\frac{1}{5^2}$152 +$\frac{1}{6^2}$162 +$\frac{1}{7^2}$172 +.....+
=> bạn biết làm rồi nên thôi
=> nói thật ra là bí
Lời giải:
$A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}$
$=\frac{1}{4}(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2})$
Để bài toán đc cm, ta cần chỉ ra: $1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 2$
Thật vậy:
$1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}$
$=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}$
$=2-\frac{1}{50}< 2$
Do đó ta có đpcm.