Question

[Trần Hoàng Minh]

Chứng minh rằng số tự nhiên \(A=1\times2\times3\times...\times69\times70\times\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{70}\right)⋮71\)

Answer 1

[Đào Trọng Luân]

Ta có:

\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{70}=\left[1+\frac{1}{70}\right]+\left[\frac{1}{2}+\frac{1}{69}\right]+\left[\frac{1}{3}+\frac{1}{68}\right]+...+\left[\frac{1}{35}+\frac{1}{36}\right]\)

\(=\frac{71}{1.70}+\frac{71}{2.69}+\frac{71}{3.68}+...+\frac{71}{35.36}\)

\(=71\left[\frac{1}{1.70}+\frac{1}{2.69}+\frac{1}{3.68}+...+\frac{1}{35.36}\right]⋮71\)

=> \(A=1\times2\times3\times4\times...\times70\times\left[1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{70}\right]⋮71\)=> ĐPCM

AI THẤY ĐÚNG NHỚ ỦNG HỘ NHA

Answer 2

[Trà My]

Xét \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{70}=\left(1+\frac{1}{70}\right)+\left(\frac{1}{2}+\frac{1}{69}\right)+...+\left(\frac{1}{35}+\frac{1}{36}\right)\)

\(=\frac{71}{1.70}+\frac{71}{2.69}+...+\frac{71}{35.36}=71\left(\frac{1}{1.70}+\frac{1}{2.69}+...+\frac{1}{35.36}\right)\)

=>\(A=1.2.3.4...71.\left(\frac{1}{1.70}+\frac{1}{2.69}+...+\frac{1}{35.36}\right)⋮71\)

Vậy A chia hết cho 71