Question

tính A=(1-1/2²) . (1-1/3²) . ........... . (1-1/2020²)

Answer 1

Xét theo thừa số tổng quát:

$1-\frac{1}{t^2}=\frac{t^2-1}{t^2}=\frac{\left(t-1\right)\left(t+1\right)}{t^2}$

Thay vào bài toán:

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

$A=\left(\frac{\left(2-1\right)\left(2+1\right)}{2^2}\right)\left(\frac{\left(3-1\right)\left(3+1\right)}{3^2}\right)\left(\frac{\left(4-1\right)\left(4+1\right)}{4^2}\right)....\left(\frac{\left(n-1\right)\left(n+1\right)}{n^2}\right)$

$A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}......\frac{\left(n-1\right)\left(n+1\right)}{n.n}$

$A=\frac{\mathrm{1.2.3.....}\left(n-1\right)}{\mathrm{2.3.4.....}n}.\frac{\mathrm{3.4.5......}\left(n+1\right)}{\mathrm{2.3.4......}n}$

$=\frac{1}{n}.\frac{n+1}{2}$

$=$

$\backslash (\backslash dfrac\{n+1\}\{2n\}\backslash )$