\(P=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{6}\right)+...+\left(1-\dfrac{1}{1225}\right)+\left(1-\dfrac{1}{1275}\right)\\ \Rightarrow\dfrac{P}{2}=\left(\dfrac{1}{2}-\dfrac{1}{6}\right)+\left(\dfrac{1}{2}-\dfrac{1}{12}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{2550}\right)\\ =\left(\dfrac{1}{2}-\dfrac{1}{2\cdot3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3\cdot4}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{50\cdot51}\right)\\ =\dfrac{1}{2}\cdot49-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\\ =\dfrac{49}{2}-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\\ =\dfrac{49}{2}-\dfrac{1}{2}+\dfrac{1}{51}=\dfrac{1225}{51}\\ \Rightarrow P=\dfrac{2450}{51}\)

\(B=-1-\frac{1}{3}-\frac{1}{6}-...-\frac{1}{1225}\)

\(=-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)

\(=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=-2\left(1-\frac{1}{50}\right)=-2\cdot\frac{49}{50}=-\frac{49}{25}\)

\(B=-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)

\(B=-2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\right)\)

\(B=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(B=-2\left(1-\frac{1}{50}\right)\)

\(B=-2\cdot\frac{49}{50}\)

\(B=-\frac{49}{25}\)

\(P=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{1225}\right)\left(1-\dfrac{1}{1275}\right)\)

\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{1224}{1225}.\dfrac{1274}{1275}\)

\(=\dfrac{2.2}{3.2}.\dfrac{5.2}{6.2}.\dfrac{9.2}{10.2}...\dfrac{1224.2}{1225.2}.\dfrac{1274.2}{1275.2}\)

\(=\dfrac{4}{9}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{2448}{2450}.\dfrac{2548}{2550}\)

\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{48.51}{49.50}.\dfrac{49.52}{50.51}\)

\(=\dfrac{1.2.3...48.49}{2.3.4...49.50}.\dfrac{4.5.6...51.52}{3.4.5...50.51}\)

\(=\dfrac{1}{50}.\dfrac{52}{3}\)

\(=\dfrac{26}{75}\).

Bạn tham khảo câu trả lời tại đây nhé:

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\(B=\left(\frac{1}{3}-1\right).\left(\frac{1}{6}-1\right).\left(\frac{1}{10}-1\right).......\left(\frac{1}{1225}-1\right)\left(\frac{1}{1275}-1\right)\)

\(B=\frac{-2}{3}.\frac{-5}{6}.\frac{-9}{10}......\frac{-1224}{1225}.\frac{-1274}{1275}\)

\(B=\frac{-4}{6}.\frac{-10}{12}.\frac{-18}{20}......\frac{-2448}{2450}.\frac{-2548}{2550}\)

\(B=\frac{-4}{2.3}.\frac{-10}{3.4}.\frac{-18}{4.5}.....\frac{-2448}{49.50}.\frac{-2548}{50.51}\)

\(\Rightarrow\)B có : ( 50 - 2 ) : 1 + 1 = 49 ( số hạng )

\(\Rightarrow B=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}........\frac{2448}{49.50}.\frac{2548}{50.51}.\left(-1\right)\)

     \(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.........\frac{48.51}{49.50}.\frac{49.52}{50.51}.\left(-1\right)\)

     \(B=\frac{\left(1.2.3...48.49\right).\left(4.5.6......51.52\right)}{\left(2.3.4......49.50\right).\left(3.4.5.....50.51\right)}.\left(-1\right)\)

     \(B=\frac{52}{50.3}.\left(-1\right)\)

      \(B=\frac{26}{75}.\left(-1\right)\)

Vậy \(B=\frac{-26}{75}\)

B

từ 1 đến 2012 có tất cả:

2012-1:1+1 = 2012 (số)

=>có: 2012:2 = 1006 (cặp)

Mà mỗi cặp bằng (-1)nên

tổng dãy số trên là: 1006 . (-1) = -1006

(1-2)+(2-3)+(3-4)+(5-6)+...+(2011-2012)

=-1+(-1)+(-1)+(-1)+...+(-1)

có tất cả các số -1 trên dãy số trên là

(2012-2);2+1=1006

vậy suy ra ; -1x1006=(-1006)

chac chan la dung