Question

metoda inductieii:

1)1*2+2*3+...+n(n+1)=$\frac{n}{3}$(n+1)(n+2)

2)($(6^{2n}-1) este divizibil la 35$

Answer 1

1) verificam P(1)=1·2=(1/3)·(1+1)(1+2) ⇒ 2=(1/3)·2·3=2 deci adevarata pt P(1)
Presupunem P(n) adevarata si demontram ca P(n+1) adevarata
P(n+1) = 1·2+2·3+....+n(n+1)+(n+1)(n+2)=P(n)+(n+1)(n+2)=(n/3)·(n+1)(n+2)+(n+1)(n+2)=(n+1)(n+2)[(n/3)+1]=(n+1)(n+2)[(n+3)/3]=[(n+1)/3](n+2)(n+3) deci P(n+1) adevarata

2) P(1)=(6²-1)=35 divizibil cu 35
Presupunem P(n) adevarata si demontram ca P(n+1) adevarata
P(n+1) = $(6^{2n+2}-1)$ = $6^{2n}*6^{2}-1=(6^{2n}-1)*6^{2}+6^{2}-1=P(n)*6^{2}+35$ unde P(n) este divizibil cu 35 + 35 este un numar divizibil cu 35 deci P(n+1) adevarata