Chứng minh A chia hết cho 101, biết A = 1.2.3.4…..99.100.(1+1/2+1/3+…+1/100) 30/09/2021 Bởi Ivy Chứng minh A chia hết cho 101, biết A = 1.2.3.4…..99.100.(1+1/2+1/3+…+1/100)
A=1.2.3…..99.100.(1+1/2+1/3+…+1/99+1/100) =1.2.3…..99.100.[(1+1/100)+(1/2+1/99)+(1/3+1/98)+…+(1/50+1/51)] =1.2.3…..99.100.(101/100+101/2.99+101/3.98+…+101/50.51) =1.2.3…..99.100.101.(1/100+1/2.99+…+1/50.51) Vì 101:101⇒A:101(đpcm) Vậy số tự nhiên A chia hết cho 101 Bình luận
Giải thích các bước giải: Ta có: $B=1+\dfrac12+\dfrac13+…+\dfrac1{100}$ $\to B=(1+\dfrac1{100})+(\dfrac12+\dfrac1{99})+(\dfrac13+\dfrac1{98})+….+(\dfrac1{50}+\dfrac1{51})$ $\to B=(1+\dfrac1{100})+(\dfrac12+\dfrac1{99})+(\dfrac13+\dfrac1{98})+….+(\dfrac1{50}+\dfrac1{51})$ $\to B=\dfrac{101}{1\cdot 100}+\dfrac{101}{2\cdot 99}+\dfrac{101}{3\cdot 98}+…+\dfrac{101}{50\cdot 51}$ $\to B=101\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+…+\dfrac{1}{50\cdot 51})$ Lại có: $A=1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot (1+\dfrac12+\dfrac13+…+\dfrac1{100})$ $\to A=1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot 101\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+…+\dfrac{1}{50\cdot 51})$ $\to A= 101\cdot (1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+…+\dfrac{1}{50\cdot 51}))$ Mà $ (1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+…+\dfrac{1}{50\cdot 51}))\in Z$ $\to A\quad\vdots\quad 101$ Bình luận
A=1.2.3…..99.100.(1+1/2+1/3+…+1/99+1/100)
=1.2.3…..99.100.[(1+1/100)+(1/2+1/99)+(1/3+1/98)+…+(1/50+1/51)]
=1.2.3…..99.100.(101/100+101/2.99+101/3.98+…+101/50.51)
=1.2.3…..99.100.101.(1/100+1/2.99+…+1/50.51)
Vì 101:101⇒A:101(đpcm)
Vậy số tự nhiên A chia hết cho 101
Giải thích các bước giải:
Ta có:
$B=1+\dfrac12+\dfrac13+…+\dfrac1{100}$
$\to B=(1+\dfrac1{100})+(\dfrac12+\dfrac1{99})+(\dfrac13+\dfrac1{98})+….+(\dfrac1{50}+\dfrac1{51})$
$\to B=(1+\dfrac1{100})+(\dfrac12+\dfrac1{99})+(\dfrac13+\dfrac1{98})+….+(\dfrac1{50}+\dfrac1{51})$
$\to B=\dfrac{101}{1\cdot 100}+\dfrac{101}{2\cdot 99}+\dfrac{101}{3\cdot 98}+…+\dfrac{101}{50\cdot 51}$
$\to B=101\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+…+\dfrac{1}{50\cdot 51})$
Lại có:
$A=1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot (1+\dfrac12+\dfrac13+…+\dfrac1{100})$
$\to A=1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot 101\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+…+\dfrac{1}{50\cdot 51})$
$\to A= 101\cdot (1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+…+\dfrac{1}{50\cdot 51}))$
Mà $ (1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+…+\dfrac{1}{50\cdot 51}))\in Z$
$\to A\quad\vdots\quad 101$