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a)Ta có:\(\dfrac{1}{b}-\dfrac{1}{b+1}=\dfrac{b+1-b}{b\left(b+1\right)}=\dfrac{1}{b^2+b}< \dfrac{1}{b^2}\)(do b>1)
\(\dfrac{1}{b-1}-\dfrac{1}{b}=\dfrac{b-b+1}{\left(b-1\right)b}=\dfrac{1}{b^2-b}>\dfrac{1}{b^2}\)(do b>1)
b)Áp dụng từ câu a
=>\(\dfrac{1}{2}-\dfrac{1}{3}< \dfrac{1}{2^2}< \dfrac{1}{1}-\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{4}< \dfrac{1}{3^2}< \dfrac{1}{2}-\dfrac{1}{3}\)
.........................
\(\dfrac{1}{9}-\dfrac{1}{10}< \dfrac{1}{9^2}< \dfrac{1}{8}-\dfrac{1}{9}\)
=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}< S< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
=>\(\dfrac{1}{2}-\dfrac{1}{10}< S< 1-\dfrac{1}{9}\)
=>\(\dfrac{2}{5}< S< \dfrac{8}{9}\)(đpcm)
1)Ta thấy: \(\dfrac{1}{n^2}=\dfrac{1}{n.n}< \dfrac{1}{\left(n-1\right)n}\)
=>A=\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}...+\dfrac{1}{50^2}< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49.50}\)
A<\(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)
Vậy A<2
2)Ta có:2S=6+3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\)
2S-S=(6+3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\))-(3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\))
=>S=6-\(\dfrac{3}{2^9}=\dfrac{6.2^9-3}{2^9}\)
Vậy S=\(\dfrac{6.2^9-3}{2^9}\)
B1: Tính nhanh:
\(E=\dfrac{-9}{10}\cdot\dfrac{5}{14}+\dfrac{1}{10}\cdot\dfrac{-9}{2}+\dfrac{1}{7}\cdot\dfrac{-9}{10}\)
\(E=\dfrac{-9}{10}\cdot\dfrac{5}{14}+\dfrac{-9}{10}\cdot\dfrac{1}{2}+\dfrac{1}{7}\cdot\dfrac{-9}{10}\)
\(E=\dfrac{-9}{10}\cdot\left(\dfrac{5}{14}+\dfrac{1}{2}+\dfrac{1}{7}\right)\)
\(E=\dfrac{-9}{10}\cdot\left(\dfrac{5}{14}+\dfrac{7}{14}+\dfrac{2}{14}\right)\)
\(E=\dfrac{-9}{10}\cdot1=\dfrac{-9}{10}\)
B2: Chứng tỏ rằng:
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}< 1\)
Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow1-\dfrac{1}{100}=\dfrac{99}{100}\)
Mà \(\dfrac{99}{100}< 1\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}< 1\)
Tick mình nha!
\(A=1.3+2.4+3.5+.............+97.99+98.100\)
\(A=\left(2-1\right)\left(2+1\right)+\left(3-1\right)\left(3+1\right)+.............+\left(99-1\right)\left(99+1\right)\)
\(A=2^2-1+3^2-1+..............+99^2-1\)
\(A=1+2^2+3^2+............+99^2-99\)
Mà :
\(1+2+2^2+...........+n^2=\dfrac{\left(n+1\right)\left(n+2\right)}{6}\)
\(\Rightarrow A=\dfrac{99\left(99+1\right)\left(99+2\right)}{6}-99=\dfrac{99.100.101}{6}-99\)
\(A=166650-99=166551\)
~ Học tốt ~
Nhanh nha