CHỨNG MINH
a, 1.3.5.7.....197.199<\(\frac{101.102.102.....200}{1+2+2^2+2^3+...+2^{99}}\)
b, \(\frac{-1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{2499}{2500}>\frac{-1}{\left(-7\right)^2}\)
CMR: 1.3.5.7......197.199=\(\frac{101}{2}+\frac{102}{2}+\frac{103}{2}+........+\frac{200}{2}\)
chứng minh rằng : 1.3.5.7....197.199 = \(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}....\frac{200}{2}\)
1.3.5.....197.199 = \(\frac{\left(1.3.5.....197.199\right)\left(2.4.6.....198.200\right)}{2.4.6......198.200}\)= \(\frac{1.2.3......199.200}{2^{100}.\left(1.2.3.....100\right)}=\frac{101.102.103......200}{2^{100}}=\frac{101}{2}.\frac{102}{2}.\frac{103}{2}.....\frac{200}{2}\)
Tính \(\frac{1}{1.3.5.7}+\frac{1}{3.5.7.9}+\frac{1}{5.7.9.11}+...+\frac{1}{\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)\left(2n+7\right)}\)
Chứng minh \(\frac{1}{1.3}+\frac{1}{1.3.5}+\frac{1}{1.3.5.7}+...+\frac{1}{1.3.5.7...\left(2n+1\right)}< \frac{1}{2}\)
\(VT< \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(2.VT< \frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{\left(2n+1\right)-\left(2n-1\right)}{\left(2n-1\right).\left(2n+1\right)}\)
\(2.VT< 1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\)
\(2.VT< 1-\frac{1}{2n+1}\Rightarrow VT< \frac{1}{2}-\frac{1}{2\left(2n+1\right)}< \frac{1}{2}\)
chứng tỏ :1.3.5.7.....197.199=101/2.102/2.103/2....200/2
\(\dfrac{101}{2}.\dfrac{102}{2}.\dfrac{103}{2}.\dfrac{104}{2}.....\dfrac{200}{2}\\ =\dfrac{101.102.103.104.....200}{2^{100}}\\ =\dfrac{\left(101.102.103.....200\right)\left(1.2.3.....100\right)}{2^{100}.\left(1.2.3.....100\right)}\\ =\dfrac{1.2.3.....200}{\left(2.1\right)\left(2.2\right)\left(2.3\right).....\left(2.100\right)}\\ =\dfrac{\left(1.3.5.....199\right)\left(2.4.6.....200\right)}{4.6.8.....200}\\ =1.3.5.7.....197.199\)
=> Điều phải chứng minh
CMR \(\frac{1.3.5.7............\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)............2n}\)=\(\frac{1}{2^n}\)
CMR : \(\frac{1.3.5.7..............\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...............2n}\) =\(\frac{1}{^{2^n}}\)
Ta có: \(\frac{1.3.5.7.....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}\)
\(=\frac{1.2.3.4..5.6...\left(2n-1\right).2n}{\left(2.4.6....2n\right)\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)
\(=\frac{1.2.3.4.5.6...\left(2n-1\right)}{2^n.1.2.3....n\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)
\(=\frac{1}{2^n}\left(đpcm\right)\)
chứng tỏ rằng:1.3.5.7...197.199=101/2.102/2.103/2...200/2
\(\frac{101}{2}\times\frac{102}{2}\times\frac{103}{2}\times...\times\frac{200}{2}\)
\(=\frac{1.2.3.....100.101.102.103.....200}{1.2.3.....100.2^{100}}\)
\(=\frac{\left(1.3.5.....199\right).\left(2.4.6.....200\right)}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(100.2\right)}\)
\(=1.3.5.....199\)
Chứng minh rằng:
\(\frac{1.3.5.7.....\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)....2n}=\frac{1}{2^n}\)
(với n ϵ N*)