[tex]p(n)=1+2+3+...+n= \frac{n(n+1)}{2} [/tex]
1. Etapa de verificare
n=1=>1=(1+1)/2=>1=1 (A)
2. Etapa de demonstrare p(k)->p(k+1)
p(k)=1+2+3+....+k=[tex] \frac{k(k+1)}{2} [/tex]
p(k+1)=1+2+3+....+k+(k+1)=[tex] \frac{(k+1)(k+2)}{2} [/tex]
p(k+1)=[tex] \frac{k(k+1)}{2} [/tex]+k+1=[tex] \frac{(k+1)(k+2)}{2} [/tex]
[tex]<=> \frac{k(k+1)+2(k+1)}{2} = \frac{(k+1)(k+2)}{2} \\ <=>\frac{(k+1)(k+2)}{2}=\frac{(k+1)(k+2)}{2} (A)[/tex]
=>(PIM) p(n) este adevărat pt. orice n natural nenul.
2. [tex]1+5+9+....+(4n-3)=n(2n-1)[/tex]
1. Etapa de verificare
n=1=> 1=1•1<=>1=1 (A)
2. Etapa de demonstrare p(k)->p(k+1), k≥1 fixat
p(k)=1+5+9+....+(4k-3)=k(2k-1)
p(k+1)=1+5+9+....+(4k-3)+(4k+1)=(k+1)(2k+1)
p(k+1)=k(2k-1)+(4k+1)=(k+1)(2k+1)
<=>2k^2-k+4k+1=(k+1)(2k+1)
<=>(k+1)(2k+1)=(k+1)(2k+1) (A)
=>(PIM) p(n) este adevărat pt. orice n natural nenul
