Question

9. a) va rog prin metoda inductiei matematice, Dau multe puncte!!

Answer 1

2^n≥2n-1

n=2

2^2≥2*2-1

4≥4-1

4>3 adevarat

Presupunem  Pn    adevarata.Verificam    daca   si   P(n+1)   adevarata

Pn;2^n≥2n-1

P(n+1):2^(n+1)≥2(n+1)-1

2*2^n≥2n+2-1

2*2^n≥2n+1   Adui    si    scazi    -1

2^n+2^n≥2n-1+2

2^n≥2^n-1   conf   Pn

2^n≥2  Evident

Deci    inegalitatea     adevarata   Pn=>P(n+1)

Answer 2

pt n=2

2²≥2*2-1

4≥4-1 adevarat

presupunem adevaeratPn

adica 2^n≥2n-1

atunci 2^(n+1) =2*2^n≥2(2n-1)=4n-2

dar 4n-2>2n+1=2(n+1)-1

ptc ca 4n-2n>2+1

adica 2n>3 , adevarat pt n≥2

deci 2^(n+1)>4n-2>2n+1=2(n+1)-1

Deci  Pn⇒P(n+1) propozitia este adevarata ∀n∈N,n≥2