Question

Mate clada a IXa 1+1/2+1/4+...+1/2la puterea n+1=2-1/2la putetea n+1 calculati P(k+1) inductie matematica

Answer 1

[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)
$\frac{-2+1}{2^(k+2)}= \frac{-1}{2^(k+2)} \ \textless \ =\ \textgreater \ \frac{-1}{2^(k+2)} = \frac{-1}{2^(k+2)}$  (Adevarat)
Conform principiului inductiei matematice P(n) este adevarata.