Question

1+3+5+...+(2n-1)=n^2​

Answer 1

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