Question

Cho biểu thức P = (1/2^2 - 1) x (1/3^2 - 1) x (1/4^2 - 1) . . . (1/2020^2 - 1). So sánh P với 1/2

Answer 1

Lời giải:
$P=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.....\frac{1-2020^2}{2020^2}$
$=\frac{(1-2^2)(1-3^2)...(1-2020^2)}{2^2.3^2....2020^2}$

$=\frac{(2^2-1)(3^2-1)....(2020^2-1)}{2^2.3^2...2020^2}$

$=\frac{(2-1)(2+1)(3-1)(3+1)....(2020-1)(2020+1)}{2^2.3^2...2020^2}$

$=\frac{(2-1)(3-1)....(2020-1)}{2.3....2020}.\frac{(2+1)(3+1)...(2020+1)}{2.3....2020}$

$=\frac{1.2.3...2019}{2.3....2020}.\frac{3.4....2021}{2.3...2020}$

$=\frac{1}{2020}.\frac{2021}{2}=\frac{2021}{4040}> \frac{2020}{4040}=\frac{1}{2}$

Vậy $P> \frac{1}{2}$