Trä Loi Bhangra nhé

S = 2/1×2 + 2/2×3 + 2/3×4 + 2/4×5 + ... + 2/101×102

B = 2 × (1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/101×102)

B = 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/101 - 1/102)

B = 2 × (1 - 1/102)

B = 2 × 101/102

B = 101/51

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{100.101}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{100}-\frac{1}{101}\right)\)

\(=2.\left(1-\frac{1}{101}\right)\)

\(=2.\frac{100}{101}=\frac{200}{101}\)

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 \(M=\frac{103}{303}\)

A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102

A=1-1/102=102/102-1/102=101/102

ý b thì chờ mình tí tìm cách lập luận đã nhé

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)

\(A=1-\frac{1}{102}\)

\(A=\frac{101}{102}\)

B=1/1.2-1/2.3-1/3.4-1/4.5-.......1/100.101-1/101.102

B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......+1/100-1/101+1/101-1/102

B=1-1/102

1+2.( 1/2-1/3+1/3-1/4+....+1/(x-1)-1/x+1)=3/2

1+2.(1/2-1/x+1)=3/2

1-2/x+1=3/2-1

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