Question

[Shiragami Yamato]

Chứng minh: \(\frac{3}{4!}+\frac{3}{5!}+\frac{3}{6!}+...+\frac{3}{100!}\)<\(\frac{1}{3!}\)

Answer 1

[Phùng Minh Quân]

Đặt \(A=\frac{4}{4!}-\frac{1}{4!}+\frac{5}{5!}-\frac{2}{5!}+\frac{6}{6!}-\frac{3}{6!}+...+\frac{100}{100!}-\frac{97}{100!}\) ta có : 

\(A=\frac{1}{3!}-\frac{1}{4!}+\frac{1}{4!}-\frac{2}{5!}+\frac{1}{5!}-\frac{3}{6!}+...+\frac{1}{99!}-\frac{97}{100!}\)

\(A=\frac{1}{3!}-\frac{1}{5!}-\frac{2}{6!}-...-\frac{95}{99!}-\frac{97}{100!}\)

\(A=\frac{1}{3!}-\left(\frac{1}{5!}+\frac{2}{6!}+...+\frac{95}{99!}+\frac{97}{100!}\right)< \frac{1}{3!}\) ( đpcm ) 

Vậy \(A< \frac{1}{3!}\)

Chúc bạn học tốt ~ 

Answer 2

[Shiragami Yamato]

thanks