(1+1/3)*(1+1/8)*(1+1/15)*.....*(1+1/9999)
giúp mik với=))
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VN
6
LA
27 tháng 6 2017
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+....+\frac{1}{9999}\)
=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=\(1-\frac{1}{101}=\frac{100}{101}\)
DN
0
HH
4
7 tháng 8 2017
ta có:
\(A=\left(1+\frac{1}{3}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right)....\left(1+\frac{1}{9999}\right)\)
\(A=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{10000}{9999}=\frac{2^2}{1.3}.\frac{3^2}{2.4}....\frac{100^2}{99.101}\)
\(A=\frac{\left(2.3.4.5....100\right)}{1.2.3.4....99}.\frac{\left(2.3.4...100\right)}{3.4.5..101}\)
\(A=\frac{100}{1}.\frac{2}{101}=\frac{200}{101}< \frac{202}{101}=2\)
\(\Rightarrow A< 2\)
TT
0
TD
4
27 tháng 6 2016
(1 + 1/3) × (1 + 1/8) × (1 + 1/15) × ... × (1 + 1/9999)
= 4/3 × 9/8 × 16/15 × ... × 10000/9999
= 2.2/1.3 × 3.3/2.4 × 4.4/3.5 × ... × 100.100/99.101
= 2.3.4...100/1.2.3...99 × 2.3.4...100/3.4.5...101
= 100 × 2/101
= 200/101
Ủng hộ mk nha ♡_♡
A = ( 1 + \(\dfrac{1}{3}\))\(\times\)( 1+ \(\dfrac{1}{8}\))\(\times\)( 1 + \(\dfrac{1}{15}\))\(\times\)...\(\times\)(1+\(\dfrac{1}{9999}\))
A = \(\dfrac{3+1}{3}\)\(\times\)\(\dfrac{8+1}{8}\)\(\times\)\(\dfrac{15+1}{15}\)\(\times\)...\(\times\)\(\dfrac{9999+1}{9999}\)
A = \(\dfrac{4}{3}\)\(\times\)\(\dfrac{9}{8}\)\(\times\)\(\dfrac{16}{15}\)\(\times\)\(\dfrac{10000}{9999}\)
A = \(\dfrac{2\times2}{1\times3}\)\(\times\dfrac{3\times3}{2\times4}\)\(\times\)\(\dfrac{4\times4}{3\times5}\)\(\times\)...\(\times\)\(\dfrac{100\times100}{99\times101}\)
A =\(\dfrac{2\times2\times\left(3\times4\times..\times99\right)\times\left(3\times4\times..99\right)\times100\times100}{1\times2\times\left(3\times4\times..99\right)\times\left(3\times4\times..99\right)\times100\times101}\)
A = \(\dfrac{200}{101}\)