Question

S=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{49}\)+\(\dfrac{1}{50}\)

P=\(\dfrac{1}{49}\)+\(\dfrac{2}{48}\)+\(\dfrac{3}{47}\)+...+\(\dfrac{48}{2}\)+\(\dfrac{49}{1}\)

Tính \(\dfrac{S}{P}\)

Answer 1

Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)

\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)

\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(\Rightarrow\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)