Câu hỏi của TORO ZANE

Question

A=$\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right).....\left(\dfrac{1}{100^2}-1\right)$

So sánh A voi $\dfrac{1}{2}$

Mashiro Shiina · Accepted Answer

$A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right).....\left(\dfrac{1}{100^2}-1\right)$

$A=-\dfrac{3}{4}.-\dfrac{8}{9}.-\dfrac{15}{16}.....-\dfrac{9999}{10000}$

$A=\dfrac{-\left(1.3\right)}{2.2}.\dfrac{-\left(2.4\right)}{3.3}.\dfrac{-\left(3.5\right)}{4.4}......\dfrac{-\left(99.101\right)}{100.100}$

Từ $-1$ đến $-99$ có: $\left(99-1\right):1+1=99$

Cộng thêm số $-101$ tất cả có $100$ số hạng

$A=\dfrac{1.3.2.4.3.5....99.101}{2.2.3.3.4.4.....100.100}$

$A=\dfrac{1.2.3....99}{2.3.4....100}.\dfrac{3.4.5....101}{2.3.4....100}$

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

$A=\dfrac{101}{200}$

Câu trả lời: $A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right).....\left(\dfrac{1}{100^2}-1\right)$