(1+1/1x3)x(1+1/2x4)x(1+1/3x5)x...x(1/2019x2021)
dấu trong bài là dấu nhân chứ không phải chữ
A=(1+1/1x3)x(1+1/2x4)x(1+1/3x5)x...x(1+1/99x101)
Lời giải:
Xét thừa số tổng quát $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$
Khi đó:
$1+\frac{1}{1.3}=\frac{2^2}{1.3}$
$1+\frac{1}{2.4}=\frac{3^2}{2.4}$
.........
$1+\frac{1}{99.101}=\frac{100^2}{99.101}$
Khi đó:
$A=\frac{2^2.3^2.4^2......100^2}{(1.3).(2.4).(3.5)....(99.101)}$
$=\frac{(2.3.4...100)(2.3.4...100)}{(1.2.3...99)(3.4.5...101)}$
$=\frac{2.3.4...100}{1.2.3..99}.\frac{2.3.4...100}{3.4.5..101}$
$=100.\frac{2}{101}=\frac{200}{101}$
(1+1/1x3)x(1+1/2x4)x(1+1/3x5)x...x(1+1/99x101)
\(\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99.101}\right)\)
\(=\left(\frac{3}{3}+\frac{1}{3}\right)\times\left(\frac{8}{8}+\frac{1}{8}\right)\times\left(\frac{15}{15}+\frac{1}{15}\right)\times...\times\left(\frac{9999}{9999}+\frac{1}{9999}\right)\)
\(=\frac{4}{3}\times\frac{9}{8}\times\frac{16}{15}\times...\times\frac{10000}{9999}\)
\(=\frac{4\times9\times16\times...\times10000}{3\times8\times15\times...\times9999}\)
\(=\frac{2\times2\times3\times3\times4\times4\times...\times100\times100}{1\times3\times2\times4\times3\times5\times...\times99\times101}\)
\(=\frac{2\times100}{101}=\frac{200}{101}\)
(1+1/1x3) x (1+1/2x4) x (1+1/3x5) x...x (1+1/99x101) x X = 100/101
A= (1+1/1x3)x(1+1/2x4)x(1+1/3x5)x...x(1+98/100)
(1+1/1x3)x(1+1/2x4)x(1+1/3x5)x(1+1/4x6)x(1+1/5x7)
tinh A =(1+1/1x3) x(1+1/2x4)x(1+1/3x5)x...(1+1/99x100)
Ps cuối hình như có vấn đề..........
C=(1+1/1x3)x(1+1/2x4)x(1+1/3x5)............(1+1/2014x2016)
A=(1+1/1x3)x(1+1/2x4)x(1+1/3x5)x...x(1+1/99x101)
A=1/2x(1+1/1x3)x(1+1/2x4)x(1+1/3x5)...(1+1/2021x2023)
giúp mình với