Ta có:
\(\dfrac{51}{2}\cdot\dfrac{52}{2}\cdot...\cdot\dfrac{100}{2}\\ =\dfrac{51\cdot52\cdot...\cdot100}{2^{50}}\\ =\dfrac{\left(1\cdot2\cdot...\cdot50\right)\left(51\cdot52\cdot...\cdot100\right)}{\left(1\cdot2\cdot...\cdot50\right)\cdot2^{50}}\\ =\dfrac{1\cdot2\cdot3\cdot...\cdot100}{2\cdot4\cdot6\cdot...\cdot100}\\ =1\cdot3\cdot5\cdot...\cdot99\)
Tính B-C , biết B = 1.3.5. ... .99 và C = \(\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}.......\dfrac{100}{2}\) . giúp mk nhanh nha
So sánh A và B :
\(A=1.3.5.7.....99\)
\(B=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}.....\dfrac{100}{2}\)
Cho \(S=\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{99}+\dfrac{1}{100}\)
So sánh S với\(\dfrac{1}{2}\)
SO SÁNH A VÀ B
A=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B=\(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\)
Cmr : \(\dfrac{1}{3}\) - \(\dfrac{2}{3^2}\) +\(\dfrac{3}{3^3}\) - \(\dfrac{4}{3^4}\) + ...+\(\dfrac{99}{3^{99}}\) - \(\dfrac{100}{3^{100}}\)< \(\dfrac{3}{16}\)
CMR \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
CMR 1.3.5...19=\(\dfrac{11}{2}.\dfrac{12}{2}.\dfrac{13}{2}...\dfrac{20}{2}\)
Chứng minh rằng : A = \(\dfrac{1}{2}-\dfrac{2}{2^2}+\dfrac{3}{2^3}-\dfrac{4}{2^4}+....+\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}< \dfrac{2}{9}\)
rút gọn
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}}{\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}}\)
