\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{100.101}=?\)
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PV
4
3 tháng 5 2018
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{100^2}{100.101}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}...\frac{100.100}{100.101}\)
\(=\frac{1.1.2.2.3.3...100.100}{1.2.2.3.3.4...100.101}\)
\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3....100\right).\left(2.3.4...101\right)}\)
\(=\frac{1.1}{1.101}\)
\(=\frac{1}{101}\)
WH
3 tháng 5 2018
\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}.....\frac{100^2}{100\cdot101}\)
\(=\frac{1.1}{1\cdot2}\cdot\frac{2.2}{2.3}\cdot\frac{3.3}{3.4}.....\frac{100.100}{100.101}\)
\(=\frac{\left(1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot\cdot100\right)\left(1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot101\right)}\)
\(=\frac{1}{101}\)
S
1
2 tháng 8 2019
\(\frac{1.1}{1.2}.\frac{2.2}{2.3}\frac{3.3}{3.4}...\frac{100.100}{100.101}\)
\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3...100\right).\left(2.3...101\right)}\)
\(=\frac{1}{1.101}\)
\(=\frac{1}{101}\)
k cho mk nha
LK
2
11 tháng 4 2017
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.......\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1.2.3.....100}{1.2.3....100}.\frac{1.2.3....100}{2.3.4...101}\)
\(=1.\frac{1}{101}=\frac{1}{101}\)
11 tháng 4 2017
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}\)
\(=\frac{1.2.3...99.100}{2.3.4...100.101}\)
\(=\frac{1}{101}\)
PG
0
3 tháng 4 2018
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.............\frac{100^2}{100.101}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}..........\frac{100.100}{100.101}\)
\(=\frac{\left(1.2.3............100\right).\left(1.2.3..........100\right)}{\left(1.2.3..........100\right)\left(2.3.4...........101\right)}\)
\(=\frac{1}{101}\)
TM
20 tháng 8 2017
Tử số = \(1.2.4+2.3.5+3.4.6+...+100.101.103\)
\(=1.2.\left(3+1\right)+2.3.\left(4+1\right)+3.4.\left(5+1\right)+...+100.101.\left(102+1\right)\)
\(=1.2.3+1.2+2.3.4+2.3+3.4.5+3.4+...+100.101.102+100.101\)
\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)+\left(1.2+2.3+3.4+...+100.101\right)\)
Mẫu số = \(1.2^2+2.3^2+3.4^2+...+100.101^2\)
\(=1.2.\left(3-1\right)+2.3.\left(4-1\right)+3.4.\left(5-1\right)+...+100.101.\left(102-1\right)\)
\(=1.2.3-1.2+2.3.4-2.3+3.4.5-3.4+...+100.101.102-100.101\)
\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)-\left(1.2+2.3+3.4+...+100.101\right)\)
đặt \(A=1.2.3+2.3.4+3.4.5+...+100.101.102\) và \(B=1.2+2.3+3.4+...+100.101\)
bạn tự tính : \(A=\frac{100.101.102.103}{4}=25.101.102.103\); \(B=\frac{100.101.102}{3}=100.101.34\)
rồi thay vào tìm P=\(\frac{A+B}{A-B}\)
=2(1/1.2+1/2.3+...+1/100.101)
=2(1/1-1/2+1/2-...+1/100-1/101)
=2(1-1/101)
=2.100/101
=200/101
2/1.2+2/2.3+2/3.4+...+2/100.101
= 2(2/1.2+2/2.3+2/3.4+...+2/100.101)
= 2(1/1.2+1/2.3+1/3.4+...+1/100.101)
= 2(1/1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)
= 2.(1/1-1/101)
= 2.100/101
= 200/101
Cho mình 1 đ-ú-n-g nha bạn