[tex]1+ \frac{1}{2} + \frac{1}{4} +...+ \frac{1}{2^(n+1)} =2- \frac{1}{2^(n+1)}
[/tex]
Etapa verificarii: P(1): 1+1/2+1/2^2=2-1/2^2 => 7/4=7/4 (adevarat)
Etapa 2 .Presupunem ca P(k) e adv.si demonstram P(k+1)
P(k):1+1/2+1/4+..+[tex] \frac{1}{2^(k+1)}+ \frac{1}{2^(k+2)} =2- \frac{1}{2^(k+2)}\ \textless \ =\ \textgreater \
2- \frac{1}{2^(k+1)} + \frac{1}{2^(k+2)} =2- \frac{1}{2^(k+2)}
[/tex]
2 cu 2 se reduc.
[tex]- \frac{1}{2^(k+1)} + \frac{1}{2^(k+2)} = -\frac{1}{2^(k+2)}
\ \textless \ =\ \textgreater \ -\frac{1}{2^k*2^1} + \frac{1}{2^k*2^2} =- \frac{1}{2^k*2^2}~
[/tex]
Aducem la numitorul comun: 2^(k+2)
[tex] \frac{-2+1}{2^(k+2)}= \frac{-1}{2^(k+2)} \ \textless \ =\ \textgreater \ \frac{-1}{2^(k+2)} = \frac{-1}{2^(k+2)} [/tex] (Adevarat)
Conform principiului inductiei matematice P(n) este adevarata.
