Question

inductia matematica pt. 7+10+13+...+3n+4=n(3n+11)/2
Ms!!!!!!!!!!!!!!

Answer 1

Salut !

Fie propozitia p(n): 7+10+13+....+3n+4 = n(3n+11)/2  pt ∀ n∈N*

-Etape-

I:Verificare : pentru n=1 avem  3+4=(3+11)/2 ⇔7=14/2 - A (adevarat)

II:Demonstrare : presupunem ca p(k)-A , k∈N*

Demonstram ca p(k+1) → p(k) ,adica p(k+1)-A

Avem:

p(k) = 7+10+....+3k+4 = k·(3k+11)/2

p(k+1) = 7+10+...+3k+4 + 3(k+1)+4 = (k+1)·[3(k+1)+11]/2

7+10+...+3k+4 + 3(k+1)+4 = (k+1)·[3k+3+11]/2

|_________|

        ║

       p(k)         + 3k+3+4 = (k+1)·[3k+14]/2

k·(3k+11)/2       + 3k+7 = (3k²+3k+14k+14)/2

(3k²+11k)/2       + 3k+7 = (3k²+17k+14)/2

(3k²+11k)/2       + (6k+14)/2 = (3k²+17k+14)/2 | · 2

3k²+11k            + 6k+14       = 3k²+17k+14

3k²+17k+14 = 3k²+17k+14   | - 3k²-17k-14

               0 = 0  - A

                  ↓

p(k+1) → p(k)  → Presupunerea facuta este adevarata

                                         ↓

7+10+13+...+3n+4=n(3n+11)/2 ∀ n ∈N*