Question

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A= 1/3×5 + 7/5×12 + 9/4×39

B= (1/2 - 1) × ( 1/3 - 1 ) × .....× (1/2002 - 1) × ( 1/2003 - 1)

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Answer 1

\(A=\dfrac{1}{3}\cdot5+\dfrac{7}{5}\cdot12+\dfrac{9}{4}\cdot39\)

\(A=\dfrac{5}{3}+\dfrac{84}{5}+\dfrac{351}{4}\)

\(A=\dfrac{27}{15}+\dfrac{351}{4}\)

\(A=\dfrac{6373}{60}\)

\(B=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)\cdot....\cdot\left(\dfrac{1}{2003}-1\right)\)

\(B=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot....\cdot\dfrac{-2002}{2003}\)

\(B=\dfrac{\left(-1\right)\cdot\left(-2\right)\cdot\left(-3\right)\cdot....\cdot\left(-2002\right)}{2\cdot3\cdot4\cdot...\cdot2003}\)

\(B=\dfrac{-1}{2003}\)