Question

Answer 1

\(P=\left(\dfrac{2^3+1}{2^3-1}\right)\left(\dfrac{3^3+1}{3^3-1}\right)\cdot...\cdot\left(\dfrac{20^3+1}{20^3-1}\right)\)

\(=\dfrac{\left(2+1\right)\left(2^2-2+1\right)}{\left(2-1\right)\left(2^2+2+1\right)}\cdot\dfrac{\left(3+1\right)\left(3^2-3+1\right)}{\left(3-1\right)\left(3^2+3+1\right)}\cdot...\cdot\dfrac{\left(20+1\right)\left(20^2-20+1\right)}{\left(20-1\right)\left(20^2+20+1\right)}\)

\(=\dfrac{3}{1}\cdot\dfrac{4}{2}\cdot...\cdot\dfrac{21}{19}\cdot\dfrac{3}{7}\cdot\dfrac{7}{13}\cdot\dfrac{13}{21}\cdot...\cdot\dfrac{381}{421}\)

\(=\dfrac{20\cdot21}{1}\cdot\dfrac{3}{421}=\dfrac{420\cdot3}{421}=\dfrac{1260}{421}\)