Question

Sa se demonstreze ca pentru orice n apartine N/{0}, [1/2]+[2/2]+[3/2]+...+[n/2]=[n/2]*[(n+1)/2]

Answer 1

Se  face  prin  inductie:
[1/2]=0... [(n+1)/2]=[n/2]+[1/2]=[n/2]
P(1)=[1/2]=[1/2]*[2/2] 
0=0
Presupui  Pn  adevarata.  P(n)=>P(n+1)?
P(n)=[1/2]+[2/2]+...+[n/2]=[n/2]*[(n+1)/2]
P(n+1)=[1/2]+...+[n/2]+[(n+1)/2]=[(n+1)/2]*[n+2/2]
[n/2]*[(n+1)/2]+[(n+1)/2]=[(n+1)/2]*[n+2)/2]
[n/2]*([n/2}+[1/2]+[n/2]+[1/2]=([n/2]+[1/2})*([n/2)+1)
[n/2]²+[n/2][n/2]²+1
Adevarat