1/1*2+1/2*3+...+1/n*(n+1)=
1/1*2=1-1/2
1/2*3=1/2-1/3
.............................
1/n*(n+1)=1/n-1/(n+1) Se aduna termen cu termen , se reduc termenii opusi si se obtine
=1-1/(n+1)
Demonstratie prin inductie
n=1
1/1*2=1-1/(1+1)=1/2
Presupui Pn adevarata .Verifici daca Pn+1 adevarata
Pn=1/1*2+1/2*3+...+1/n*(n+1)=1-1/(n+1)
Pn+1=1/1*2+...+1/n*(n+1)+1/(n+1)(n+2)=1-1/(n+2)
1-1/(n+1)+1/(n+1)*(n+2)=
1+[-(n+2)+1]/(n+1)*(n+2)=1-1/(n+2)
1+[-n-2+1]/(n+1)(n+2)=1-1/(n+2)
1+(-n-1)/(n+1)(n+2)=1-1/(n+2)
1-1/(n+2)=1-1/(n+2) evident Pn=>Pn+1
