Question

Chứng minh rằng  A = 1/2! + 1/3! + 1/4! + 1/5!+...+1/100! < 1

Answer 1

\(A=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....+\frac{1}{100!}\)

\(=\frac{2-1}{2!}+\frac{3-2}{3!}+....+\frac{100-99}{100!}\)

\(=\frac{2}{2!}-\frac{1}{2!}+\frac{1}{2!}-\frac{2}{3!}+....+\frac{100}{100!}-\frac{99}{100!}\)

\(=\frac{2}{2!}-\frac{99}{100!}=1-\frac{99}{100!}< 1\)

\(\Rightarrow A< 1\) (DPCM)

Answer 2

Đinh Đức Hùng sướng

Được làm CTV

Tui còn chả được

Answer 3

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

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

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

\(A=\frac{2}{2!}-\frac{99}{100!}\)= 1 - \(\frac{99}{100!}< 1\)

Vậy A < 1