Question

\(A=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{127}\). Chứng minh A>4

Answer 1

\(A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{127}\)

   \(=\left(\dfrac{1}{127}+1\right).127\div2\)

   \(=\dfrac{128}{127}.127\div2\)

   \(=128\div2\)

   = 64

Vậy A>4

Answer 2

hmmm :)

Answer 3

=1+1/2+1/3+...+1/127

   =(1/127+1).1/27÷2

   =128/127.127÷2

   =128÷2

   = 64

Vậy A>4