CHỨNG MINH
a, 1.3.5.7.....197.199<\(\frac{101.102.102.....200}{1+2+2^2+2^3+...+2^{99}}\)
b, \(\frac{-1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{2499}{2500}>\frac{-1}{\left(-7\right)^2}\)
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D
3 tháng 5 2015
1.3.5.....197.199 = \(\frac{\left(1.3.5.....197.199\right)\left(2.4.6.....198.200\right)}{2.4.6......198.200}\)= \(\frac{1.2.3......199.200}{2^{100}.\left(1.2.3.....100\right)}=\frac{101.102.103......200}{2^{100}}=\frac{101}{2}.\frac{102}{2}.\frac{103}{2}.....\frac{200}{2}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
26 tháng 9 2018
\(VT< \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(2.VT< \frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{\left(2n+1\right)-\left(2n-1\right)}{\left(2n-1\right).\left(2n+1\right)}\)
\(2.VT< 1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\)
\(2.VT< 1-\frac{1}{2n+1}\Rightarrow VT< \frac{1}{2}-\frac{1}{2\left(2n+1\right)}< \frac{1}{2}\)
0
1
19 tháng 3 2018
\(\dfrac{101}{2}.\dfrac{102}{2}.\dfrac{103}{2}.\dfrac{104}{2}.....\dfrac{200}{2}\\ =\dfrac{101.102.103.104.....200}{2^{100}}\\ =\dfrac{\left(101.102.103.....200\right)\left(1.2.3.....100\right)}{2^{100}.\left(1.2.3.....100\right)}\\ =\dfrac{1.2.3.....200}{\left(2.1\right)\left(2.2\right)\left(2.3\right).....\left(2.100\right)}\\ =\dfrac{\left(1.3.5.....199\right)\left(2.4.6.....200\right)}{4.6.8.....200}\\ =1.3.5.7.....197.199\)
=> Điều phải chứng minh
S
18 tháng 3 2017
Ta có: \(\frac{1.3.5.7.....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}\)
\(=\frac{1.2.3.4..5.6...\left(2n-1\right).2n}{\left(2.4.6....2n\right)\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)
\(=\frac{1.2.3.4.5.6...\left(2n-1\right)}{2^n.1.2.3....n\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)
\(=\frac{1}{2^n}\left(đpcm\right)\)
HD
1
DD
Đoàn Đức Hà
Giáo viên
29 tháng 5 2021
\(\frac{101}{2}\times\frac{102}{2}\times\frac{103}{2}\times...\times\frac{200}{2}\)
\(=\frac{1.2.3.....100.101.102.103.....200}{1.2.3.....100.2^{100}}\)
\(=\frac{\left(1.3.5.....199\right).\left(2.4.6.....200\right)}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(100.2\right)}\)
\(=1.3.5.....199\)
a) dat A=1+2+22+23+...+299
2.A=2+22+23+24+...+2100
2.A-A= 2+23+24+...+2100-(1+2+22+23+...+299)
A=2100-1
----> 1.3.5.7...197.199<\(\frac{101.102.103....200}{2^{100}-1}\)
Dat B =1.3.5.7...197.199
B=\(\frac{1.3.5.7....197.199...2.4.6.8....200}{2.4.6.8....200}\)
B= \(\frac{1.2.3.4.5....199.200}{2.4.6.8....200}\)
B=\(\frac{1.2.3.4.5......199.200}{2^{100}.\left(1.2.3.4...100\right)}\) ( tu 2 den 200 co 100 so hang nen duoc 2100)
B =\(\frac{101.102.103....200}{2^{100}}\)
---->\(\frac{101.102.103....200}{2^{100}}
b> A= \(\frac{1.3.5.7....2499}{2.4.6.8....2500}\) chon B=\(\frac{2.4.6.8...2500}{3.5.7.9...2501}\)
A.B = \(\frac{1.3.5.7....2499.2.4.6.8...2500}{2.4.6.8...2500.3.5.7.....2499.2501}=\frac{1}{2501}\)
Nhan xet
\(\frac{1}{2}+\frac{1}{2}=1\)
\(\frac{2}{3}+\frac{1}{3}=1\)
vi 1/2 >1/3----> 1/2 <2/3
cm tuong tu ta se co A<B
---> A.A<A.B
---->A2<A.B
===> A2 <\(\frac{1}{2501}