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\(=\dfrac{1\cdot2+2\cdot3+3\cdot4+4\cdot5+...+49\cdot50}{2}\)
\(=\dfrac{49\cdot50\cdot51}{3}=49\cdot50\cdot17=41650\)
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\(P=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{6}\right)+...+\left(1-\dfrac{1}{1225}\right)+\left(1-\dfrac{1}{1275}\right)\\ \Rightarrow\dfrac{P}{2}=\left(\dfrac{1}{2}-\dfrac{1}{6}\right)+\left(\dfrac{1}{2}-\dfrac{1}{12}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{2550}\right)\\ =\left(\dfrac{1}{2}-\dfrac{1}{2\cdot3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3\cdot4}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{50\cdot51}\right)\\ =\dfrac{1}{2}\cdot49-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\\ =\dfrac{49}{2}-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\\ =\dfrac{49}{2}-\dfrac{1}{2}+\dfrac{1}{51}=\dfrac{1225}{51}\\ \Rightarrow P=\dfrac{2450}{51}\)
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\(P=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{1225}\right)\left(1-\dfrac{1}{1275}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{1224}{1225}.\dfrac{1274}{1275}\)
\(=\dfrac{2.2}{3.2}.\dfrac{5.2}{6.2}.\dfrac{9.2}{10.2}...\dfrac{1224.2}{1225.2}.\dfrac{1274.2}{1275.2}\)
\(=\dfrac{4}{9}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{2448}{2450}.\dfrac{2548}{2550}\)
\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{48.51}{49.50}.\dfrac{49.52}{50.51}\)
\(=\dfrac{1.2.3...48.49}{2.3.4...49.50}.\dfrac{4.5.6...51.52}{3.4.5...50.51}\)
\(=\dfrac{1}{50}.\dfrac{52}{3}\)
\(=\dfrac{26}{75}\).
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\(2C=2+6+12+20+...+2450\)
\(2C=1.2+2.3+3.4+4.5+...+49.50\)
\(6C=1.2.3+2.3.3+3.4.3+...+49.50.3\)\(6C=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(6C=1.2.3+2.3.4-1.2.3+...+49.50.51-49.50.51\)
\(6C=49.50.51\)
\(6C=124950\)
\(C=20825\)
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2xC=2+6+12+20+...+2450
2xC=1x2+2x3+3x4+4x5+...+49x50
6xC=1x2x3+2x3x3+3x4x3+...+49x50x3
6xC=1x2x3+2x3x(4-1)+3x4x(5-2)+...+49x50x(51-48)
6xC=1x2x3+2x3x4-1x2x3+....+49x50x51-48x49x50
6xC=49x50x51
6xC=124950
C=20825
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