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\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}+\frac{1}{50}\)
\(=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)
\(=\frac{2}{\left(2+1\right).2}+\frac{2}{\left(3+1\right).3}+...+\frac{2}{\left(99+1\right).99}+\frac{1}{50}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2.\left(\frac{50}{100}-\frac{1}{100}\right)+\frac{1}{50}=2.\frac{49}{100}+\frac{1}{50}\)
\(=\frac{49}{50}+\frac{1}{50}=1\)
\(S=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{99}{49^2\cdot50^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+.....+\frac{1}{49^2}-\frac{1}{50^2}\)
\(=1-\frac{1}{50^2}=\frac{2499}{2500}\)
\(T=\frac{1}{\left(2-1\right)\left(2+1\right)}+\frac{1}{\left(3-1\right)\left(3+1\right)}+...+\frac{1}{\left(50-1\right)\left(50+1\right)}\)
\(=\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{49\cdot51}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\cdot\left(1+\frac{1}{2}-\frac{1}{51}\right)=\frac{151}{204}\)
Vì \(\frac{2499}{2500}>\frac{151}{204}\)nên S>T
JOKER_Võ Văn Quốc, T = \(\frac{1}{2}.\left(1-\frac{1}{51}+\frac{1}{2}-\frac{1}{50}\right)\)mới đúng
Sẽ dễ hơn nếu bạn chia ra 2 vế \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)và \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{48+50}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+99}+\frac{1}{50}\)
\(=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+\frac{1}{\frac{\left(4+1\right).4}{2}}+....+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)
\(=\frac{1}{\frac{3.2}{2}}+\frac{1}{\frac{4.3}{2}}+\frac{1}{\frac{5.4}{2}}+....+\frac{1}{\frac{100.99}{2}}+\frac{1}{50}\)
\(=\frac{2}{3.2}+\frac{2}{4.3}+\frac{2}{5.4}+...+\frac{2}{100.99}+\frac{1}{50}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2.\frac{49}{100}+\frac{1}{50}=\frac{49}{50}+\frac{1}{50}=1\)
\(A=\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+...+\frac{2}{99}-\frac{2}{100}+\frac{1}{50}\)
\(A=\frac{2}{2}-\frac{2}{100}+\frac{1}{50}=1\)
A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}-\dfrac{1}{5}\)+ \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)
A = 1