Question

[Namlun_A8]

A=151/102 - 1/6 - 1/12 - 1/20 -...- 1/2550

Answer 1

[Phùng Minh Quân]

Ta có : 

\(A=\frac{151}{102}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-...-\frac{1}{2550}\)

\(A=\frac{151}{102}-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\right)\)

\(A=\frac{151}{102}-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)\)

\(A=\frac{151}{102}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(A=\frac{151}{102}-\left(\frac{1}{2}-\frac{1}{51}\right)\)

\(A=\frac{151}{102}-\frac{49}{102}\)

\(A=\frac{151-49}{102}\)

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

\(A=1\)

Vậy \(A=1\)

Chúc bạn học tốt ~ 

Answer 2

[Emma Granger]

\(A=\frac{151}{102}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-...-\frac{1}{2550}\)

\(A=\frac{151}{102}-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{50\cdot51}\right)\)

\(A=\frac{151}{102}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(A=\frac{151}{102}-\left(\frac{1}{2}-\frac{1}{51}\right)\)

\(A=\frac{151}{102}-\frac{49}{102}=1\)