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\(d=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).........\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{3}.\frac{9}{2.4}.............\frac{10000}{99.101}\)
\(=\frac{2.2}{3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}............\frac{100.100}{99.101}\)
\(=\frac{2.3.4..........100}{2.3.4............99}.\frac{2.3.4...........100}{3.4...........101}\)
\(=100.\frac{2}{101}\)\(=\frac{200}{101}\)
\(C=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{1993}{1994}\)
\(=\frac{1\times2\times3\times...\times1993}{2\times3\times4\times...\times1994}\)
\(=\frac{1}{1994}\) (Giản ước còn lại như này)
Từ đề bài suy ra: A=3/4 x 8/9 x ...x 9800/9801 x 9999/10000
=>A=<1x3/2x2> x <2x4/3x3> x ... x <99x101/100x100>
=>A=(1x2x...x99)/(2x3x...x100) x (3x4x...x101)/(2x3x...x100)
=>A=1/100 x 101/2 = 101/200
a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)
a) =3/2 . 4/3 . 5/4 ...100/99
=\(\frac{3.4.5...100}{2.3.4..99}\)
=\(\frac{100}{2}\)
b) =
a, Đúng rồi đó
b, \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)+....+\left(1+\frac{1}{39.41}\right)\)
= \(\frac{4}{1.3}.\frac{9}{2.4}.....\frac{1600}{39.41}\)
= \(\frac{2.2.3.3....40.40}{1.3.2.4....39.41}\)
= \(\frac{\left(2.3....40\right)\left(2.3....40\right)}{\left(1.2....39\right)\left(3.4....41\right)}\)
= \(\frac{40.2}{41}\)
= \(\frac{80}{41}\)
\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{8^2}\right)\)
\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{8^2-1}{8^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.....\frac{7.9}{8.8}\)
\(=\frac{\left(1.2.....7\right).\left(3.4.....9\right)}{\left(2.3.....8\right).\left(2.3.....8\right)}\)
\(=\frac{1.9}{8.2}=\frac{9}{16}\)
101/180