Question

1 .thục hiện phép tính

\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\)

Answer 1

Đặt :

\(S=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+..................+\dfrac{99}{100!}\)

\(\Rightarrow S=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+................+\dfrac{100-1}{100!}\)

\(\Rightarrow S=\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{1!}+\dfrac{4}{4!}-\dfrac{1}{1!}+..................+\dfrac{100}{100!}-\dfrac{1}{100!}\)

\(\Rightarrow S=1-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+................+\dfrac{1}{99!}-\dfrac{1}{100!}\)

\(\Rightarrow S=1-\dfrac{1}{100!}< 1\)

\(\Rightarrow S< 1\)

Vậy \(\dfrac{1}{2!}+\dfrac{2}{3!}+..................+\dfrac{99}{100!}< 1\rightarrowđpcm\)