Question

[34 Tạ Khánh Ngọc]

tính bằng cách thuận tiện : (1 - 1 phần 4) nhân (1 - 1 phần 9) nhân ( 1 - 1 phần 16 ) nhân ... nhân (1- 1 phần 576) nhân (1 - 1 phần 625}

Answer 1

[Nguyễn Việt Lâm]

Ta có:

\(1-\dfrac{1}{n^2}=\dfrac{n^2-1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)

Áp dụng:

\(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{625}\right)\)

\(=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{25^2}\right)\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{24.26}{25^2}\)

\(=\dfrac{1.2.3...24}{2.3.4...25}.\dfrac{3.4.5...26}{2.3.4...25}=\dfrac{1}{25}.\dfrac{26}{2}=\dfrac{13}{25}\)

Answer 2

[Trần Ngọc Vân]

\(=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{575}{576}.\dfrac{624}{625}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{24.26}{25.25}\)

\(=\dfrac{\left(1.2...24\right).\left(3.4...26\right)}{\left(2.3...25\right).\left(2.3...25\right)}\)

\(=\dfrac{1.26}{25.2}=\dfrac{1.2.13}{25.2}=\dfrac{13}{25}\)