Cho M= (3/1.4 + 3/2.6 + 3/3.8 + 3/4.10 + ... + 3/49.100)/ (1-1/4)(1-1/5)(1-1/6)...(1-1/100) . Chứng tỏ rằng M có giá trị là một số nguyên.

* Hướng dẫn giải

Đặt $\mathrm{A}=\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+\frac{3}{4.10}+\mathrm{...}+\frac{3}{49.100}$ ;

$\mathrm{B}=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{6}\right)\mathrm{...}\left(1-\frac{1}{100}\right)$;

Ta có $\mathrm{A}=\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+\frac{3}{4.10}+\mathrm{...}+\frac{3}{49.100}$

$\begin{array}{l}=\frac{6}{2.4}+\frac{4}{4.6}+\frac{6}{6.8}+\frac{6}{8.10}+\mathrm{...}+\frac{6}{98.100}\\ =3.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\mathrm{...}+\frac{2}{98.100}\right)\\ =3.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\mathrm{...}+\frac{1}{98}-\frac{1}{100}\right)\\ =3.\left(\frac{1}{2}-\frac{1}{100}\right)\\ =3.\frac{49}{100}\end{array}$.

Với: $\mathrm{B}=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{6}\right)\mathrm{...}\left(1-\frac{1}{100}\right)$

$=\frac{3}{4}.\frac{4}{5}.\frac{5}{6}\mathrm{...}\frac{99}{100}$.

Suy ra $\mathrm{M}=\frac{\mathrm{A}}{\mathrm{B}}=3.\frac{49}{100}:\frac{3}{100}=\frac{3.49}{100}.\frac{100}{3}=49$

Ta thấy 49 $\in$ ℕ.

Do đó biểu thức M là một giá trị nguyên.