Các câu hỏi tương tự
Chứng minh rằng:\(\frac{51}{2}+\frac{52}{2}+...+\frac{100}{2}=1.3.5...99\)
Chứng tỏ rằng:
1.3.5...99=\(\frac{51}{2}.\frac{52}{2}...\frac{100}{2}\)
Chứng tỏ rằng: \(1.3.5...99=\frac{51}{2}.\frac{52}{2}...\frac{100}{2}\)
Chứng minh rằng :
\(\frac{51}{2}.\frac{52}{2}.....\frac{100}{2}=1.3.5....99\)
CM : \(1.3.5.....99=\frac{51}{2}.\frac{52}{2}.....\frac{100}{2}\)
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) \(\frac{51}{2}+\frac{52}{2}+...+\frac{100}{2}=1.3.5...99\)
So sánh: 1.3.5....99 với \(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}....\frac{100}{2}\)
Chứng tỏ rằng:
1.3.5. ... .99=\(\frac{51}{2}\).\(\frac{52}{2}\).\(\frac{53}{2}\). ... .\(\frac{100}{2}\)
chứng minh rằng:\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)