tinh a: [1+1/1.3].[1+1/2.4].....[1+1/2013.2014]
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1
N
1
30 tháng 10 2018
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....+\left(1+\frac{1}{99.101}\right)\)
\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)
\(=\frac{2.3.4.....100}{1.2.3.....99}.\frac{2.3.4.....100}{3.4.5.....101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
KC
0
NN
tinh: (1+\(\frac{1}{1.3}\))(1+\(\frac{1}{2.4}\))(1+\(\frac{1}{3.5}\)).......(1+\(\frac{1}{99.101}\))
1
19 tháng 8 2018
Ta có: \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{9999}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.....\frac{10000}{9999}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.....\frac{100.100}{99.101}\)
\(=\frac{2.2.3.3.4.4.....100.100}{1.3.2.4.3.5.....99.101}\)
\(=\frac{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}\)
\(=\frac{2.3.4.....100}{1.2.3.....99}.\frac{2.3.4.....100}{3.4.5.....101}\)
\(=100.\frac{2}{101}\)
\(=\frac{200}{101}\)
Ta có :\(\left(1+\frac{1}{1.3}\right)+\left(1+\frac{1}{2.4}\right)+...+\left(1+\frac{1}{18.20}\right)\)
= \(\frac{4}{1.3}+\frac{9}{2.4}+...+\frac{361}{18.20}\)
= \(\frac{2.2.3.3.4.4.....18.18.19.19}{1.3.2.4.3.5.....17.19.18.20}\)
= \(\frac{2.19}{1.20}=\frac{19}{10}\)