1/ Cho \(A=\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)
Chứng minh rằng: \(A=\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\frac{ }{ }\right)\right]\)
2/ Tính \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2005^2}\). Chứng minh \(A< 1\)
3/ Cho \(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Chứng minh: \(\frac{1}{2}< A< 1\)
GIÚP MÌNH NHA, MÌNH ĐANG CẦN GẤP.MÌNH SẼ TICK AI NHANH NHẤT!!
1.Cho A =\(\frac{1}{1.102}\)+\(\frac{1}{2.103}\)+.....+\(\frac{1}{299.400}\)
Chứng minh rằng:
A=\(\frac{1}{101}\)\([\)\((\)1+\(\frac{1}{2}\)+......+\(\frac{1}{101}\)\()\)- \((\)\(\frac{1}{300}\)+\(\frac{1}{301}\)+......+\(\frac{1}{400})\)\(]\)
1/ Cho \(A=\frac{1}{1.102}+\frac{1}{2.203}+...+\frac{1}{299.400}\)
Chứng minh rằng: \(A=\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)
2/ Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2005^2}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}\)
a) So sánh \(A\)với \(B\) b) Chứng minh: \(A< 1\)
3/ Cho \(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Chứng minh: \(\frac{1}{2}< A< 1\)
GIÚP MÌNH NHA, MÌNH CẢM ƠN. MÌNH ĐANG CẦN GẤP!!!!
Tính bằng cách hợp lí:
a) A=\(\left(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}\right):\left(\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\right)\)
b) B=\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{200}\right):\left(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+....\frac{198}{2}+\frac{199}{1}\right)\)
\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)chứng minh rằng \(\frac{1}{299}\left(\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right)\)=\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)
1/ Tính bằng cách thuận tiện nhất:
\(A=\)\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+....+\frac{9899}{9900}\)
2/ Cho \(A=\)\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}\)
Chứng minh rằng: \(A=\)\(\frac{1}{299}\).\(\left[\left(1+\frac{1}{2}+....+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+....+\frac{1}{400}\right)\right]\)
GIÚP MÌNH VS, MÌNH ĐANG CẦN GẤP.MÌNH SẼ TICK CHO AI NHANH NHẤT!!!!!
Cho A=\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}\)
CMR:A=\(\frac{1}{299}.\left(1+\frac{1}{2}+.......+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+.....+\frac{1}{400}\right)\)
ở chỗ 1/299 là nhân với ngoặc vuông nha bạn nào giải hộ mình rthì i li-ke
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+\frac{1}{5.995}+......+\frac{1}{999.1}}\)
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+......+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+4.5+......+98.99}\)
\(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+......+\frac{1}{100.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+......+\frac{1}{299.400}}\)
\(D=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{97}-......-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+......+\frac{1}{500}}\)
Bài 1
a rút gọn B=\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
b Chứng minh A=\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}>\frac{5}{8}\)