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\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{49.51}{50^2}\)
\(=\frac{2.3....49}{3.4....50}.\frac{4.5....51}{3.4....50}\)
\(=\frac{2}{50}.\frac{17}{1}\)
\(=\frac{17}{25}\)
Ta có : \(A=\frac{8}{9}.\frac{15}{16}.....\frac{2499}{2500}\)
\(A=\frac{8.15.....2499}{9.16.....2500}\)
\(A=\frac{\left(2.4\right).\left(3.5\right).....\left(49.51\right)}{\left(3.3\right).\left(4.4\right).....\left(50.50\right)}\)
\(A=\frac{\left(2.3....49\right).\left(4.5....51\right)}{\left(3.4....50\right).\left(3.4.....50\right)}\)
\(A=\frac{2\left(3.4.....49\right).\left(4.5.....50\right).51}{\left(3.4.....49\right).50.3.\left(4.5.....50\right)}\)
\(A=\frac{2.51}{3.50}\)
\(A=\frac{2.17.3}{3.25.2}\)
\(A=\frac{17}{25}\)
Do S = \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}\)
\(\Rightarrow\)S = \(\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+...+\left(1-\frac{1}{50^2}\right)\)
\(\Rightarrow\)S=(1+1+1+...+1) - \(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
\(\Rightarrow\)S=49-\(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Dễ thấy:\(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)không phải là số tự nhiên
\(\Rightarrow\)S\(\notin N\)
\(B=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\)
\(=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{50^2}\)
\(=\left(1+1+1+...+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
\(=49.1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}=\dfrac{49}{50}< 1\)
\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>-1\)
\(\Rightarrow B=49.1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>49-1=48\)
\(\Rightarrow\) B > 48 (đpcm)
