c)a1=0
a2=1
a3=3/2
a4=7/4
a5=15/8
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an=[2^(n-1)-1]/2^(n-2) (A
Se demonstreaza prin inductie,Se considera an adevarata .Se verifica daca si an+1=adevarata
an+1=[2^n-1]/2^(n-1) relatia (B
Dar an+1 = prin definitie=1/2*an+1
an+1=1/2*[2^(n-1)-1]/2^(n-2)+1=[2^(n-1)-1]/2^(n-1)=[2^(n-1)-1+2^(n-1)]/2^(n-1)=(2*^2^(n-1)-1)/2^(n-1)=(2^(n)-1)/2^(n-1)=(B
Deci Pn=>Pn+1
SE va studia an dpdv al monotoniei Se compara an cu an+1\
Pt aceasta se amplifica cu 2 an si se obtine
2^(n-1)-2]/2^(n-1)<2^(n)/2^(n-1) an<an=1 => an sir crescator
,Marginirea an>0
an=[2^(n-2)-1]/2^(n-2)=[2^(n-2)/2^(n-2)]-1/2^(n-2)=2-1/2^(n-2)<2 C)
deci an∈[0,2) margoinit
an =marginit si monoton=>convergent
Vom trece la limita in relatia c)
lim an=lim(2-1/2^(n-2))=2
