Question

Valoarea numarului N $N = \dfrac{8000}{19}\cdot( \dfrac{1}{1+2+3....+100} +\dfrac{1}{1+2+3....+101} +......+\dfrac{1}{1+2+3....+1999})$ este egala cu

Answer 1

Răspuns:

Explicație pas cu pas:

1/(1+2+3+.....+100)=1/100×101/2=2/100×101

1/(1+2+3+.....+101)=1/101×102/2=2/101×102

1/(1+2+3+.......+1999)=1/1999×2000/2=2/1999×2000

1/100×101=1/100-1/101

1/101×102=1/101-1/102

....................................

1/1999×2000=1/1999-1/2000

       S=2(1/100-1/2000)=2(20-1)/2000=2×19/2000

N=8000/19×2×19/2000=8

N=8