chứng minh rằng : 1.3.5.7....197.199 = \(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}....\frac{200}{2}\)
Chứng tỏ rằng :1.5.7...197.199=\(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}...\frac{200}{2}\)
Chứng tỏ rằng: 1.3.5.7.9. ... .197.199 = 101/2 . 102/2 . 103/2 . ... . 200/2
Giúp mình nhé các bạn!
CMR: 1.3.5.7......197.199=\(\frac{101}{2}+\frac{102}{2}+\frac{103}{2}+........+\frac{200}{2}\)
Chứng minh rằng:
a) \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)=\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
b) \(\frac{51}{2}+\frac{52}{2}+...+\frac{100}{2}=1.3.5...99\)
Chứng tỏ rằng \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....+\frac{1}{200}>\frac{1}{2}\)
Chi tiết rõ ràng nha
Chứng minh rằng: \(C=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{7}{12}\)
Tính: \(\frac{101}{2}.\frac{102}{6}.\frac{103}{10}...\frac{200}{398}\)
Chứng tỏ rằng : \(\frac{1}{101}\)+ \(\frac{1}{102}\)+ \(\frac{1}{103}\)+ .......+ \(\frac{1}{200}\)>\(\frac{1}{2}\)