Câu hỏi của Nguyễn Huy Trường Lưu

Question

Tính bằng cách thuận thiện:

$\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)...\left(1+\dfrac{1}{99\cdot101}\right)$

Nguyễn Đức Trí · Accepted Answer

$\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{99.101}\right)$

$=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}....\dfrac{100^2}{99.101}$

$=\dfrac{2.3.4...100}{1.2.3.4...99}.\dfrac{2.3.4...100}{3.4.5....101}$

$=\dfrac{100}{1}.\dfrac{2}{101}$

$=\dfrac{200}{101}$Câu trả lời: $\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{99.101}\right)$

$=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}....\dfrac{100^2}{99.101}$

$=\dfrac{2.3.4...100}{1.2.3.4...99}.\dfrac{2.3.4...100}{3.4.5....101}$

$=\dfrac{100}{1}.\dfrac{2}{101}$

$=\dfrac{200}{101}$