a) = 1-1/2+1/2-1/3+...+1/99 -1/100 =1-1/100=99/100=0,99
Observa, terog, ca toti termenii se reduc in afara de primul si ultimul.
b) S = ∑1/k(k+2)=1/2∑(1/k - 1/(k+2)), cu k=1,2,3,...,98 si dezvoltam suma, dandu-i valori lui k:
S=1/2(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/95-1/97+1/96-1/98+1/97-1/99+1/98-1/100) = 1/2(1+1/2-1/99-1/100) = 1/2(1.1/2 -199/ 99x100)
Succes in continuare!
