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\(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
Nên B<\(\dfrac{1}{4}\)
B=\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\)
Nên B>\(\dfrac{1}{6}\)
Dat A=/32+1/42+1/52+1/62+...+1/1002<1/2.3+1/3.4+1/4.5+1/5.6+...+1/99.100 A<1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/99-1/100<1/2 Chung to...
=> 1/32+1/42+1/52+ ....+ 1/1002<1/2.3+1/3.4+1/4.5+...+1/99.100
=> 1/32+1/42+1/52+ ....+ 1/1002<1/2-1/100=49/100<1/2
=> 1/32+1/42+1/52+ ....+ 1/1002<1/2 (đpcm)
( k cho mình nha )
Đặt \(B=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(\Rightarrow B< \dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\)
\(\Rightarrow B< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow B< \dfrac{1}{2}-\dfrac{1}{100}\left(1\right)\)
mà \(\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\left(2\right)\)
\(\left(1\right),\left(2\right)\rightarrow B< \dfrac{1}{2}\left(đpcm\right)\)
Ta có:
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{100}\)
Mà \(\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)
=> \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\left(đpcm\right)\)
\(\frac{1}{3^2}< \frac{1}{2.3}\); \(\frac{1}{4^2}< \frac{1}{3.4}\); \(\frac{1}{5^2}< \frac{1}{4.5}\); ......; \(\frac{1}{100^2}< \frac{1}{99.100}\)
=> \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.....+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
Lại có: \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{2}-\frac{1}{100}=\frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)
Vậy: \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.....+\frac{1}{100^2}< \frac{1}{2}\)=> đpcm
đặt A=1/3²+1/4²+1/5²+……1/100²
B=1/2.3+1/3.4+...+1/99.100
=1/2-1/3...+1/99-1/100
=1/2-1/100<1/2 (1)
mà A=1/3²+1/4²+1/5²+……1/100²<B=1/2.3+1/3.4+...+1/99.100 (2)
kết hợp từ (1),(2)ta được A<B<1/2
=>A<1/2
Chỉ cần một bước so sánh đơn giản là em có thể đưa bài toán này về dạng quen thuộc nhé :)
\(\frac{1}{3^2}<\frac{1}{2.3},\frac{1}{4^2}<\frac{1}{3.4},...,\frac{1}{100^2}<\frac{1}{99.100}.\)
Như vậy, ta được \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100}<\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}<\frac{1}{2}\)
Chúc em luôn học tập tốt :)