Question

sa se calculeze: S=
$\frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + ...... + \frac{1}{1 + 2 + 3 + ...... + 2014}$

Answer 1

folosim urmatoarea strategie
stim ca 1+2+3+   +n= n*(n+1)/2
deci un termen k al sumei noastre va fi
ak=1/(1+2+3+...+k)=1/[k*(k+1)/2]=2/[k*(k+1)]
dar se demonstreaza usor ca 1/[k*(k+1)=1/k-1/(k+1) (aduci la acelasi numitor)
atunci
ak=2/k-2/(k+1). Dand valori lui k >=2 obtin

S=(2/2-2/3) + (2/3-2/4) +(2/4-2/5)+.......+(2/2014 - 2/2015) dupa scaderi succesive
S=2/2-2/2015=(2*2015-4) / 4030=4026/4030