Question

1, Tính tổng sau

a, A=4/3x7 + 4/7x11 + 4/11x15 + ........+4/100x104

Answer 1

\(A=\frac{4}{3X7}+\frac{4}{7X11}+\frac{4}{11X15}+...+\frac{4}{100X104}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)

\(=\frac{1}{3}-\frac{1}{104}\)

\(=\frac{101}{312}\)

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Answer 2

\(A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{100.104}\)

\(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)

\(A=\frac{1}{3}-\frac{1}{104}=\frac{104}{312}-\frac{3}{312}=\frac{101}{312}\)

Answer 3

A=4/3.7 +4/7.11 +4/11.15 +...+4/100.104 

A=1/3 −1/7 +1/7 −1/11 +1/11 −1/15 +...+1/100 −1/104 

A=1/3 −1/104 =104/312 −3/312 =101/312 

Answer 4

Ta có : \(A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+......+\frac{4}{100.104}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{100}-\frac{1}{104}\)

\(=\frac{1}{3}-\frac{1}{104}\)

\(=\frac{101}{312}\)