aratati ca 1+2+3+...+n=[n×(n+1)]:2 pentru orice nr natural n

[tex]P(n):\quad 1+2+3+...+n = \dfrac{n(n+1)}{2} \\ \\P(1):\quad1 = \dfrac{1(1+1)}{2} \quad (A)\\ P(2):\quad 1+2 = \dfrac{2(2+1)}{2} \quad (A) \\ \\ P(k):\quad 1+2+3+...+k= \dfrac{k(k+1)}{2}\\\\\begin{array}{lcl}P(k+1):& &1+2+3+..+(k+1) = \dfrac{(k+1)(k+2)}{2} \\ \\ & &(1+2+3+...+k)+(k+1) = \dfrac{(k+1)(k+2)}{2}\\ \\ & &\dfrac{k(k+1)}{2}+k+1 =\dfrac{(k+1)(k+2)}{2} \\ \\ & & \dfrac{k(k+1)+2(k+1)}{2} = \dfrac{(k+1)(k+2)}{2}\\ \\& & \dfrac{(k+1)(k+2)}{2} = \dfrac{(k+1)(k+2)}{2}\quad (A) \end{array}[/tex]

Am demonstrat că relația e adevărată pentru orice numar natural n prin inductie matematică.

Explicație pas cu pas:

Pn=1+2+3+...+n=n(n+1)/2

n=1,P1:1=1(1+1)/2⇒1=1-Adev

n=2;P2:1+2=2(2+1)/2⇒3=3-Adve

n=3;P3:1+2+3=3(3+1)/2⇒6=6-Adev

n=k;Pk:1+2+3+...+k=k(k+1)/2-Adev

n=k+1;P(k+1):1+2+3+...+(k+1)=(k+1)(k+2)/2 ?? (O demonstram)

1+2+3+...+k+(k+1)=k(k+1)/2+(k+1)⇒(k(k+1)+2(k+1))/2=

=(k+1)(k+1)/2-Deci si P(k+1) este adevarat;

Raspuns:∀ n∈N* ⇒Pn-Adevarat

Bafta!