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Ta có nếu theo quy luật như trên thì sẽ có 1 thừa số là\(\frac{1}{49}-\frac{1}{7^2}\)
Mà chúng bằng 0 nên tích trên bằng 0
\(\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)
\(=1+1+...+1+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)(có 49 số 1)
\(=\left(1+\frac{48}{2}\right)+\left(1+\frac{47}{3}\right)+...+\left(1+\frac{2}{48}\right)+\left(1+\frac{1}{49}\right)+1\)
\(=\frac{50}{2}+\frac{50}{3}+...+\frac{50}{48}+\frac{50}{49}+\frac{50}{50}\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)\)
Chúc bạn học tốt.
\(A< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{49.50.51}.\)
\(2A< \frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{49.50.51}\)
\(2A< \frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{51-49}{49.50.51}\)
\(2A< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{49.50}-\frac{1}{50.51}\)
\(2A< \frac{1}{2}-\frac{1}{50.51}< \frac{1}{2}\Rightarrow A< \frac{1}{4}< \frac{1}{2}\)
2A=1+1/2+.........+1/2^51
2A-A={1+1/2+......+1/2^51}-{1/2+1/2^2+.....+1/2^50}
A=1-1/2^50
=>A<1
\(A=\frac{1}{2}+\frac{1}{2^2}+\cdot\cdot\cdot+\frac{1}{2^{50}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{2^{49}}\)
\(\Rightarrow2A-A=\left(1+\cdot\cdot\cdot+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\cdot\cdot\cdot\cdot+\frac{1}{2^{50}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{50}}\)
\(\Rightarrow A\) < \(1\)
Ta có : \(\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)\left(\frac{1}{49}-\frac{1}{4^2}\right).......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)
\(=\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)......\left(\frac{1}{49}-\frac{1}{7^2}\right)......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)
\(=\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)......0......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)
\(=0\)