đề hơi sai, mk sửa lại nhé
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(\Leftrightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(\Leftrightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(\Leftrightarrow2A=\frac{1}{1.2}-\frac{1}{99.100}\)
\(\Leftrightarrow2A=\frac{4950}{9900}-\frac{1}{9900}\)
\(\Leftrightarrow2A=\frac{4949}{9900}\)
\(\Leftrightarrow A=\frac{4949}{9900}\div2\)
\(\Leftrightarrow A=\frac{9898}{9900}=\frac{4949}{4950}\)
\(A=2+22+...+222222222...2222\left(\text{50 chữ số 2}\right)\)
\(\Leftrightarrow A=2\left(1+11+...+111111111...1111\right)\)
\(\Leftrightarrow A=\frac{2}{9}\left(9+99+...+9999999999...99\right)\)
\(\Leftrightarrow A=\frac{2}{9}\left[\left(10-1\right)+\left(10^2-1\right)+...+\left(10^{50}-1\right)\right]\)
\(\Leftrightarrow A=\frac{2}{9}\left[\left(10+10^2+...+10^{50}\right)-50\right]\)
Đặt \(B=\left[\left(10+10^2+...+10^{50}\right)\right]\)
\(\Leftrightarrow10B=\left[\left(10^2+10^3+...+10^{51}\right)\right]\)
\(\Leftrightarrow10B-B=\left[\left(10^2+10^3+...+10^{51}\right)\right]-10-10^2-...-10^{50}\)
\(\Leftrightarrow10B-B=10^{51}-10\)
\(\Rightarrow A=\frac{2}{9}\left(10^{51}-10-50\right)\)
\(\Rightarrow A=\frac{2}{9}\left(10^{51}-60\right)\)
