Question

Demonstrati ca pentru orice numar natural nenul n are loc egalitatea

1+2+3+...+n(n+1):2
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Answer 1

Răspuns:

Se folosește metoda inducției matematice

P(1): 1=1(1+1)/2 => 1=1*2/2 =>1 =1 (A)

P(k): 1+2+3+…+k= k(k+1)/2= S

P(k+1): 1+2+3+…+k+(k+1)=(k+1)(k+2)/2=S'

S'=S+k+1

k(k+1)/2+k+1=(k+1)(k+2)/2  | *2

k(k+1)+2k+2=(k+1)(k+2)

k²+k+2k+2=k²+k+2k+k+2 (A)

=> P(n) (A) ∀ n∈N*