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Ta có: \(\frac{1.3.5.7.....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}\)
\(=\frac{1.2.3.4..5.6...\left(2n-1\right).2n}{\left(2.4.6....2n\right)\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)
\(=\frac{1.2.3.4.5.6...\left(2n-1\right)}{2^n.1.2.3....n\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)
\(=\frac{1}{2^n}\left(đpcm\right)\)
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 =\(\frac{101.102.103....200}{2^{100}}\)
---->\(\frac{101.102.103....200}{2^{100}}<\frac{101.102.103....200}{2^{100}-1}\)
ta co : 2100 >2100-1
--->\(\frac{1}{2^{100}}<\frac{1}{2^{100}-1}\)
---> \(\frac{101.102.103...200}{2^{100}}<\frac{101.102.103...200}{2^{100}-1}\)
----> dpcm
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 <\(\frac{1}{2501}<\frac{1}{2500}=\frac{1}{50^2}\)
==> A2<1/502
--> A <1/50
ma 1/50<1/49
nen A<1/49
--> A < 1/72
---> A. (-1) >(-1).1/72
---> -A>-1/72
\(J=\frac{1}{6}\left(\frac{1}{1.3.5}-\frac{1}{3.5.7}+\frac{1}{3.5.7}-\frac{1}{5.7.9}+...+\frac{1}{19.21.23}-\frac{1}{21.23.25}\right)\)
\(=\frac{1}{6}\left(\frac{1}{1.3.5}-\frac{1}{21.23.25}\right)\)
Đến đây em tự tính nhé, số hơi to thì phải :))
Chúc học tốt!!!!!
\(J=\frac{1}{6}\left(\frac{1}{1.3.5}-\frac{1}{3.5.7}+\frac{1}{3.5.7}-\frac{1}{5.7.9}+...+\frac{1}{19.21.23}-\frac{1}{21.23.25}\right)\)
\(=\frac{1}{6}\left(\frac{1}{1.3.5}-\frac{1}{21.23.25}\right)\)
Đến đây em tự tính nhé, số hơi to thì phải :))
Chúc học tốt!!!!!