Cho A= 1/2 . 3/4 . 5/6 .... 2499/2500
Chứng minh A<1/49
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TN
14 tháng 8 2015
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}}
TN
14 tháng 8 2015
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}
YA
22 tháng 2 2017
a) Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
. . .
\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2^2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\right)\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(1+1-\frac{1}{50}\right)\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\cdot\frac{99}{50}=\frac{99}{200}< \frac{100}{200}=\frac{1}{2}\left(đpcm\right)\)
b) Ta có :
\(B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)
\(\Rightarrow1-\frac{1}{4}+1-\frac{1}{9}+...+1-\frac{1}{2500}>48\)
\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< 49\)
Lại có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
. . .
\(\frac{1}{50^2}< \frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{50^2}< \frac{49}{50}< 1\)
\(\Rightarrow-\left(\frac{1}{2^2}+...=\frac{1}{50^2}\right)>1\)
\(\Rightarrow49-\left(\frac{1}{2^2}+...+\frac{1}{50^2}\right)>49-1=48\)
hay \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}>48\left(đpcm\right)\)
6 tháng 5 2017
a) gọi \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
gọi \(B=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+1-\frac{1}{50}\)
\(=2-\frac{1}{50}< 2\)
\(\Rightarrow A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\)
b) Ta thấy \(\frac{1}{37}< \frac{1}{35}< \frac{1}{31}< \frac{1}{30}\), \(\frac{1}{61}< \frac{1}{53}< \frac{1}{47}< \frac{1}{45}\)
Do đó : \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{30}.3+\frac{1}{45}.3=\frac{1}{2}\)
c) \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)
\(=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
Ta thấy vế trong ngoặc nhỏ hơn 1
\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>48\)
S
30 tháng 4 2019
A = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/99*100
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
A = 1 - 1/100
A = 99/100
B = 5/1*4 + 5/4*7 + .... + 5/100*103
B = 5/3*(3/1*4 + 3/4*7 + ... + 3/100*103)
B = 5/3*(1 -1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)
B = 5/3*(1 - 1/103)
B = 5/3* 102/103
ko ai trả lời thì để mình
C/M : n/n+1 < n+1/n+2
1 - n/n+1 = 1/n+1
1 - n/n + 2 = 1/n+2
Vì 1/n+1 > 1/n+2 nên n/n+1 < n+1/n+2
1/2 . 3/4 . 5/6 ... 2499/2500 < 1/2 . 2/3 . 3/4 ... 2501/2502
=1/2501 < 1/2500 (1/50) 2
1/50 < 1/49 => A <1/49