a) Presupun P(k) adv,demonstrez P(k+1)
P(k)= 1+3+6+...+(2k-1)=k²
P(k+1)=1+3+..+[2(k+1)-1]= (k+1)² - notez cu *
=P(k)+[2(k+1)-1]
=k²+(2k+1)
=k² + 2k + 1 - notez cu **
*=** (adevarat,daca restrangem ** obtinem binomul (a+b)² )
b) Presupun P(k) adevarat,demonstrez P(k+1)
P(k)=1³+2³+...+k³=[tex] \frac{ k^{2} (k+1)^{2} }{4} [/tex]
P(k+1)= 1³+...+(k+1)³= [tex] \frac{ (k+1)^{2}[(k+1)+1] ^{2} }{4} [/tex] - *
=P(k)+(k+1)³
=[tex] \frac{k^{2}(k+1)^{2} }{4} [/tex] + (k+1)³
=[tex] \frac{ k^{2}(k^{2} +2k+1) }{4} [/tex] + k³+1³+3k(k+1) - dupa ce afli rezultatul final aducand la acelasi numitor,adunand termenii asemenea etc,il notezi cu **
ca sa demonstrezi ca *=**, la * desfaci parantezele si iti va da la fel
