Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Khi đó:
\(1-\frac{2}{2.3}=\frac{1.4}{2.3}\) ; \(1-\frac{2}{3.4}=\frac{2.5}{3.4}\) ; ... ; \(1-\frac{2}{101.102}=\frac{100.103}{101.102}\)
\(\Rightarrow M=\frac{1.4}{2.3}\cdot\frac{2.5}{3.4}\cdot\cdot\cdot\frac{100.103}{101.102}\)
\(M=\frac{\left(1.2...100\right).\left(4.5...103\right)}{\left(2.3...101\right).\left(3.4...102\right)}=\frac{103}{101.3}=\frac{103}{303}\)
Vậy \(M=\frac{103}{303}\)
Bài 15 :
a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=1-\frac{1}{2020}=\frac{2019}{2020}< \frac{2020}{2020}=1\)
b) Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)
\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\)
\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)
\(A=\frac{1}{2^{1001}}-\frac{1}{2}\)
Tới đây là so sánh đi nhé
Cái này mình làm hôm qua rồi mà '-'
a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)
\(\Rightarrow A< 1\)
b) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\)
\(2A-A=A\)
\(=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{999}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{1000}}\)
\(=1-\frac{1}{2^{1000}}\)
\(\Rightarrow A=1-\frac{1}{2^{1000}}< 1\left(đpcm\right)\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{100^2}{100.101}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}...\frac{100.100}{100.101}\)
\(=\frac{1.1.2.2.3.3...100.100}{1.2.2.3.3.4...100.101}\)
\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3....100\right).\left(2.3.4...101\right)}\)
\(=\frac{1.1}{1.101}\)
\(=\frac{1}{101}\)
\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}.....\frac{100^2}{100\cdot101}\)
\(=\frac{1.1}{1\cdot2}\cdot\frac{2.2}{2.3}\cdot\frac{3.3}{3.4}.....\frac{100.100}{100.101}\)
\(=\frac{\left(1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot\cdot100\right)\left(1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot101\right)}\)
\(=\frac{1}{101}\)
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+2015}\)
\(=\frac{2}{1.2}+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+2+3\right).3}{2}}+.....+\frac{1}{\frac{\left(2015+1\right).2015}{2}}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+....+\frac{2}{2015.2016}\)
\(\frac{1.1}{1.2}.\frac{2.2}{2.3}\frac{3.3}{3.4}...\frac{100.100}{100.101}\)
\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3...100\right).\left(2.3...101\right)}\)
\(=\frac{1}{1.101}\)
\(=\frac{1}{101}\)
k cho mk nha
\(\frac{-7}{11}.\frac{11}{19}+\frac{-7}{11}.\frac{8}{19}+\frac{-4}{11}\)
\(=\frac{-7}{11}.\left(\frac{11}{19}+\frac{8}{19}\right)+\frac{-4}{11}\)
\(=\frac{-7}{11}.1+\frac{-4}{11}\)
\(=\frac{-7}{11}+\frac{-4}{11}=\frac{-11}{11}=-1\)
~ Hok tốt ~
Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow B=1-\frac{1}{2019}\)
\(\Rightarrow B=\frac{2018}{2019}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.......\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1.2.3.....100}{1.2.3....100}.\frac{1.2.3....100}{2.3.4...101}\)
\(=1.\frac{1}{101}=\frac{1}{101}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}\)
\(=\frac{1.2.3...99.100}{2.3.4...100.101}\)
\(=\frac{1}{101}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{100^2}{100.101}\)
\(=\frac{1.1.2.2.3.3...100.100}{1.2.2.3.3.4.4...100.101}\)
\(=\frac{\left(1.2.3...100\right)\left(1.2.3...100\right)}{\left(1.2.3..100\right)\left(2.3.4...101\right)}=\frac{1}{101}\)