a, Đúng rồi đó
b, \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)+....+\left(1+\frac{1}{39.41}\right)\)
= \(\frac{4}{1.3}.\frac{9}{2.4}.....\frac{1600}{39.41}\)
= \(\frac{2.2.3.3....40.40}{1.3.2.4....39.41}\)
= \(\frac{\left(2.3....40\right)\left(2.3....40\right)}{\left(1.2....39\right)\left(3.4....41\right)}\)
= \(\frac{40.2}{41}\)
= \(\frac{80}{41}\)
\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{8^2}\right)\)
\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{8^2-1}{8^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.....\frac{7.9}{8.8}\)
\(=\frac{\left(1.2.....7\right).\left(3.4.....9\right)}{\left(2.3.....8\right).\left(2.3.....8\right)}\)
\(=\frac{1.9}{8.2}=\frac{9}{16}\)