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\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{50}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(A=B-2C\left(đpcm\right)\)
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{49}+\left(\frac{1}{5}\right)^{50}\)
\(5M=1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{48}+\left(\frac{1}{5}\right)^{49}\)
5M - M = \(1-\left(\frac{1}{5}\right)^{50}\)hay 4M = \(1-\left(\frac{1}{5}\right)^{50}\)< 1
\(\Rightarrow M=\frac{1-\left(\frac{1}{5}\right)^{50}}{4}< \frac{1}{4}\)
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{50}\)(1)
\(\Rightarrow5M=1+\frac{1}{5}+...+\left(\frac{1}{5}\right)^{49}\)(2)
Lấy (2)-(1) ta có
\(\Rightarrow4M=1-\left(\frac{1}{5}\right)^{50}\)
\(\Rightarrow M=\frac{1-\frac{1}{5^{50}}}{4}\)
Do \(1-\frac{1}{5^{50}}< 1\)
\(\Rightarrow M< \frac{1}{4}\)
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
\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)-\left(\frac{-15}{12}+\frac{6}{11}-\frac{48}{49}\right)\)
\(=\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}+\frac{15}{12}-\frac{6}{11}+\frac{48}{49}\)
\(=\left(\frac{-1}{4}-\frac{5}{3}+\frac{15}{12}\right)+\left(\frac{7}{33}-\frac{6}{11}\right)+\frac{48}{49}\)
\(=\left(\frac{-3}{12}-\frac{20}{12}+\frac{15}{12}\right)+\left(\frac{7}{33}-\frac{18}{33}\right)+\frac{48}{49}\)
\(=\left(\frac{-23}{12}+\frac{15}{12}\right)+\left(\frac{-11}{33}\right)+\frac{48}{49}\)
\(=\frac{-2}{3}+\left(\frac{-1}{3}\right)+\frac{48}{49}\)
\(=-1+\frac{48}{49}\)
\(=-\frac{1}{49}\)
Ùng hộ mk nha ^_-
\(\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)
\(=1+1+...+1+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)(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.