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Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\left(đpcm\right)\)
Ta có: \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Ta xét A= \(\frac{1}{5^2}+\frac{1}{6^2}+..+\frac{1}{100^2}\)
\(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}...+\frac{1}{100.101}\)
=> \(A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
=> \(A>\frac{1}{5}-\frac{1}{101}\)
=> \(A>\frac{96}{505}>\frac{96}{576}=\frac{1}{4}\)
Ta có : \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
=> \(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(A< \frac{1}{4}-\frac{1}{100}\)
=> \(A< \frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)
Sorry bạn nha , mình bấm nhầm nút
\(A=\frac{5}{4}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A< \frac{5}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A< \frac{5}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< \frac{5}{4}+\frac{1}{2}-\frac{1}{100}< \frac{5}{4}+\frac{1}{2}=\frac{7}{4}\)
\(\Rightarrow\)\(A< \frac{7}{4}\)
Vậy , \(\frac{5}{4}< A< \frac{7}{4}\left(ĐPCM\right)\)
BÀI KHÓ CỦA TRƯỜNG MÌNH ĐÓ THI HK2
GIÚP MÌNH NHÉ!!!!!!THANKS!!!!!!
Ta có : \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}\left(đpcm\right)\)
Chúc bạn học tốt !!!
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+.......+\frac{1}{100^2}<\frac{1}{2}\)
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+........+\frac{1}{100^2}\)<\(\frac{1}{0.2}+\frac{1}{2.4}+\frac{1}{4.6}+.......+\frac{1}{98.100}\)
\(S=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}<\frac{50}{100}=\frac{49}{100}<\frac{1}{2}\)
Vậy \(\frac{49}{100}<\frac{1}{2}\)
Ta có 1/22<1/2*3
1/42<1/3*4
. . .
1/1002<1/99*100
=> S<1/2*3+1/3*4+...+1/99*100
=> S<1/2-1/3+1/3-1/4+...+1/99-1/100
=>S<1/2-1/100
=>S<49/100
Mà 49/100<1/2
=>S<1/2
Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+....+\frac{1}{100^2}\)
Ta có: \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};\frac{1}{5^2}< \frac{1}{4.5};......;\frac{1}{100^2}< \frac{1}{99.100}\)
\(=>A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}\)
\(=>A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=>A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.....+\frac{1}{100^2}< \frac{1}{2}\left(đpcm\right)\)
Bạn xem lời giải của mình nhé:
Giải:
Gọi \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
\(\frac{1}{3^2}< \frac{1}{3.4}\\ \frac{1}{4^2}< \frac{1}{4.5}\\ ...\\ \frac{1}{100^2}< \frac{1}{99.100}\\ \Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{3}-\frac{1}{100}\\ \frac{1}{3}< \frac{1}{2}\Rightarrow\frac{1}{3}-\frac{1}{100}< \frac{1}{2}\\ \Rightarrow A< \frac{1}{2}\)
Chúc bạn học tốt!