Răspuns:
1+3+5+...+2n-1=
(2·1-1)+(2·2-1)+...+2n-1=
2·(1+2+....+n)-(1+1+...+1)=
2n·(n+1):2-n=
n·(n+1)-n=n·(n+1-1)=n·(n+0)=n·n=n²
______________
P(k): 1+3+5…+2k−1=k²
P(1): 1=1²
ecuatia este adevarata pentru n=1
P(k): 1+3+5+...+2k−1=k²
1+3+5+…+2k−1+2(k+1)−1=k²+2(k+1)−1
=>1+3+5+...+2k−1+2k+1=k²+2k+1=>1+3+5+...+2k−1+2k+1=(k+1)²
=>1+3+5+…+2k−1+2k+1=(k+1)²:P(k+1)
am demostrat ca P(1) si P(k+1) sunt adevarate =>P(k) adevarat
deci: 1+3+5+…+(2n−1)=n²
