Salut am si eu nevoie de niste ajutor la inductii.
Răspunsuri la întrebare
Răspuns:
Explicație pas cu pas:
respectam cele 3 conditii"
1. verificam adevarul identitatii pentru n=1
(2·1-1)·2·1·(2·1+1)=1·2·3 adevarat.
2. admitem ca identitatea este adevarata pentru n=k
1·2·3+3·4·5+...(2k-1)·2k·(2k+1)=k·(k+1)·(2k²+2k-1)
3. Sa verificam adevarul identitatii pentru n=k+1,
1·2·3+3·4·5+...(2k-1)·2k·(2k+1)+(2(k+1)-1)·2(k+1)(2(k+1)+1)=(k+1)(k+1+1)(2(k+1)²+2(k+1)-1)=(k+1)(k+2)(2(k+1)²+2(k+1)-1), la asta trebuie sa ajungem
1·2·3+3·4·5+...(2k-1)·2k·(2k+1)+(2(k+1)-1)·2(k+1)(2(k+1)+1)=k·(k+1)·(2k²+2k-1)+(2k+1)·2(k+1)(2k+3)=(k+1)·(k(2k²+2k-1)+(2k+1)(2k+3))=(k+1)·(2k³+2k²-k+8k²+16k+6)=(k+1)((2k³+8k²+8k)+2k²+7k+6)=(k+1)(2k(k+2)²+2k²+8k+8-k-2)=(k+1)(2k(k+2)²+2(k+2)²-(k+2))=(k+1)(k+2)(2k(k+2)+2(k+2)-1)=(k+1)(k+2)((k+2)(2k+2)-1)=(k+1)(k+2)(2(k+1+1)(k+1)-1)=(k+1)(k+2)(2(k+1)²+2(k+1)-1)
Ufff ~~~ am ajuns la ce tindeam ~~~