Question

chứng minh rằng: \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+....+\frac{99}{100!}

Answer 1

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

Answer 2

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

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

= \(1\)\(-\)\(\frac{1}{100!}\)\(< \)\(1\)\(\left(đpcm\right)\)

Answer 3

hai bạn làm đúng zòi

chúc các bạn hk tốt>.<