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\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+2\right)}\)
\(\Rightarrow\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{1.2.3}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Rightarrow k=2\)
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{99.100}\right).y=\frac{49}{100}\Leftrightarrow\left(\frac{99.50-1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{99.50-1}{99}\right).y=49\Leftrightarrow\left(99.50-1\right).y=99.49\Rightarrow y=\frac{99.49}{99.50-1}\)
Ta có 1/1.2-1/2.3=2/1.2.3;1/2.3-1/3.4=2/2.3.4 .....1/98.99-1/99.100=2/98.99.100 2A=2/1.2.3+2/2.3.4+....+2/98.99.100 = 1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100 = 1/2-1/99.100 = 4949/9900 A =4949/19800
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{100-98}{98.99.100}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{18000}\)
=1-/2-1/3+1/2-1/3-1/4+1/5-1/6-1/7+1/6-1/7-1/8-.........-1/98-1/99-1/100
=1-1/100
=99/100
Ta có : \(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\right)\)
\(\Rightarrow S=2.\left(1-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}=\frac{99}{50}\)
=2.(1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+.........+\(\frac{1}{99}\)-\(\frac{1}{100}\))
=2.(1-\(\frac{1}{100}\))
S= 2.\(\frac{99}{100}\)
S=\(\frac{99}{50}\)
Tính tổng dãy dấu ngoặc trước
Đặt \(S=1\cdot2+2\cdot3+3\cdot4+...+98\cdot99\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot(4-1)+...+98\cdot99\cdot(100-97)\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot3\cdot4+...+98\cdot99\cdot100-97\cdot98\cdot99\)
\(3S=98\cdot99\cdot100\Rightarrow S=\frac{1}{3}\cdot98\cdot99\cdot100\)
Thay vào đề bài,ta có :
\(\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=12\frac{6}{7}:\frac{-3}{2}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=12\frac{6}{7}\cdot\frac{2}{-3}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{90}{7}\cdot\frac{2}{-3}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{-30}{7}\cdot\frac{2}{-1}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{-60}{-7}=\frac{60}{7}\)
\(\Leftrightarrow\frac{1}{3}\cdot98\cdot99\cdot100\cdot x=\frac{60}{7}\cdot26950\)
\(\Leftrightarrow\frac{1}{3}\cdot98\cdot99\cdot100\cdot x=231000\)
\(\Leftrightarrow323400\cdot x=231000\)
\(\Leftrightarrow x=231000:323400=\frac{5}{7}\)
Tử thần sai từ dòng:
\(\frac{\frac{1}{3}.98.99.100.x}{26950}=\frac{30}{7}.\frac{2}{-1}\Leftrightarrow12x=-\frac{60}{7}\Leftrightarrow x=\frac{-5}{7}\)
A = 9(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\))
A = 9(\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\))
A = 9(1 - \(\frac{1}{100}\))
A = 9.\(\frac{99}{100}\)=\(\frac{891}{100}\)=8,91
Vì \(\frac{1}{1.2}=\frac{1}{2}\)
Mà \(\frac{1}{1}-\frac{1}{2}=\frac{2}{2}-\frac{1}{2}=\frac{1}{2}\)
Nên trong bài toán: \(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)
Mấy cái kia cũng vậy nên bạn yên tâm nha!!!!
A : 9 = 1/1.2 + 1/2.3 + 1/3.4 + ..... + 1/98.99 + 1/99.100
A : 9 = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/98 - 1/99 + 1/99 - 1/100
A : 9 = 1 - 1/100
A : 9 = 100/100 - 1/100
A : 9 = 99/100
A = 9 . 99/100
A = 891/100 = 8,91 = 8 91/100
