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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 : \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
=> \(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(A< \frac{1}{4}-\frac{1}{100}\)
=> \(A< \frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)
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: \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\)
b) gọi dãy là A ta có:
\(\frac{1}{2^2}\)<\(\frac{1}{1.2}\)
\(\frac{1}{3^2}\)<\(\frac{1}{2.3}\)
.
............
...........
\(\frac{1}{100^2}\)<\(\frac{1}{99.100}\)
đặt D=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+.......+\(\frac{1}{99.100}\)
D=1-1/2+1/2-1/3+.......+1/99-1/100
D=1-1/100=99/100
vì A <D => A<1
K NHA
B = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}<1\)
\(B<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(B<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}\)
\(B<1-\frac{1}{8}<1\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(B< 1-\frac{1}{8}< 1\)
Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow...............< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow...................< 1-\frac{1}{100}\)
\(\Rightarrow..............< \frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow....................< 1\)
Vậy..................
Đặt \(P=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Có \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
...........
\(\frac{1}{100^2}< \frac{1}{99.100}\)
=> \(P< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)
=> \(P< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
=> \(P< 1-\frac{1}{100}< 1\)
=> P < 1
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)(Đpcm)
Tta có
1/22 < 1/(1.2)= 1-1/2
1/32 <1/(2.3)=1/2-1/3
1/42 <1/(3.4)=1/3-1/4
......
1/1002 < 1/99-1/100
cộng vế với vế ta được 1/22 +1/32+...+1/1002< 1-1/2+1/2-1/3+....+1/99-1/100=1-1/100
=> ĐPCM