Các câu hỏi tương tự
CMR (1/7^2)+(1/7^4)+...+(1/7^ 4n-2)-(1/7^4n)+...+(1/7^98)+(1/7^100) < 1/50
CMR :
\(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)< \(\frac{1}{50}\)
CMR:1/7^2-1/7^4+...+1/74n-2-1/74n+...+1/798-1/7100<1/50
Chứng minh rằng :
1/7^2 - 1/7^4 + 1/7^6-1/7^8 +...+ 1/7^98 - 1/7 ^100 < 1/ 50
chứng minh rằng : \(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-...+\frac{1}{7^{98}}-\frac{1}{7^{100}}< \frac{1}{50}\)
CMR : \(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{4n-2}}-\frac{1}{7^n}+...+\frac{1}{7^{98}}+\frac{1}{7^{100}}< \frac{1}{50}\)
Giải hộ mình nhé
Chứng minh rằng: \(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-.....+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)\(< \frac{1}{50}\)
1/7^2 - 1/7^4+1/7^6 - 1/7^8+...+1/7^98 - 1/7^100 = ?
chứng minh rằng \(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}<\frac{1}{50}\)
ai nhanh minh k cho