chứng minh 1/2! + 2/3! + 3/4! +…..+99/100! <1 19/07/2021 Bởi Brielle chứng minh 1/2! + 2/3! + 3/4! +…..+99/100! <1
Giải thích các bước giải: $C= \dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+…+\dfrac{99}{100!}$ $= \dfrac{2-1}{2!}+\dfrac{3-1}{3!}+…+\dfrac{100-1}{100!}$ $= \dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+…+\dfrac{100}{100!}-\dfrac{1}{100!}$ $= \dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+…+\dfrac{1}{99!}-\dfrac{1}{100!}$ $= 1-\dfrac{1}{100!}< 1$ Bình luận
Giải thích các bước giải:
$C= \dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+…+\dfrac{99}{100!}$
$= \dfrac{2-1}{2!}+\dfrac{3-1}{3!}+…+\dfrac{100-1}{100!}$
$= \dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+…+\dfrac{100}{100!}-\dfrac{1}{100!}$
$= \dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+…+\dfrac{1}{99!}-\dfrac{1}{100!}$
$= 1-\dfrac{1}{100!}< 1$