Referat - Metode de raţionament

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Metode de rationament
Metoda directa( modus poneus) - pornind de la o propozitie A si folosind principiul silogismului demonstram ca o alta propozitie este adevarata.
Metoda indirecta – reducerea la absurd ce se bazeaza pe echivalenta (p(q) ( ((q((p)
Metoda inductiei
Metoda inductiei
A. Egalitati
P(n): 1+2+3+...+n=n(n+1)/2 n(1
I. P(1): 1=1 (A)
II. P(n) (A)(P(n+1) (A)
P(n+1):1+2+3...+n+(n+1)=(n+1)(n+2)/2
n(n+1)/2+(n+1)=(n+1)(n+2)/2
(n+1)(n+2)/2=(n+1)(n+2)/2 (A)
I+II( P(n) (A) (n(1
Deci: 1+2+3...+n = (k = n(n+1)/2 , n(1

P(n): 1? +2? +3? +...+n?= n(n+1)(2n+1)/6 n(1
I. P(1): 1=1 (A)
II. P(n) (A) (P(n+1) (A)
P(n+1): 12 +22 +32 +...+n2+(n+1)2=(n+1)(n+2)(2n+3)/6
n(n+1)(2n+1)/6+(n+1)? =( n+1)(n+2)(2n+3)/6
(n+1)(n+2)(2n+3)/6=( n+1)(n+2)(2n+3)/6 (A)
I+II(P(n) (A) (n(1
12 +22 +32 + ... +n2 = (k2 = n(n+1)(2n+1)/6, n(1

13+23+33+...+n3= [n(n+1)/2]2 n(1
I. P(1): 1=1 (A)
II. P(n) (A) (P(n+1) (A)
P(n+1):13+23+33+...+n3+(n+1)3 = [(n+1)(n+2)/2]2
[n(n+1)/2]2 +(n+1)3= [(n+1)(n+2)/2]2
[(n+1)(n+2)/2]2 = [(n+1)(n+2)/2]2 (A)
I+II(P(n) (A) (n(1
13+23+33+...+n3=(k3 =
[n(n+1)/2]2, n(1

(a1+a2+..+an(((a1(+(a2(+...+(an( n(2

Exemple de egalitati rezolvate prin inductie
ex 1: S=1(4+2(7+3(10+...+n(3n+1)
=(k(3k+1) = (3k2+k = (3k2+(k = 3(k2+(
= 3n(n+1)(2n+1) /6+n(n+2)/2
= n(n+1)(2n+1)+n(n+2)/2
= n(n+1)(2n+1+1)/2
= n(n+1)2
ex 2: p(n): 1(4+2(7+...n(3n+1)=n(n+1)2 n(1
I P(1): 1C4=1(1+1)2 ( 4=4 (A)
II P(n) (P(n+1):1(4+2(7+...+n(3n+1)+(n+1)(3n+4)=(n+1)(n+2)2
n(n+1)2+(n+1)(3n+4) = (n+1)(n+2)2
(n+1)(n2+n+3n+4) = (n+1)(n+2)2
I+II(P(n) (A) (n(1
ex 3: p(n): 1/(1(3)+1/(3(5)+...+1/[(2n-1)(2n+1)]=n/(2n+1)
1/[(2k-1)(2k+1)]=1/2 [1/(2k-1)-1/(2k+1)] ((1/[(2k-1)(2k+1)]=n/(2n+1)
S1=1/(1(3)=1/3
S2=S1+1/(1(3)=2/5
S3=S2+1/(5(7)=3/7
I P(1): 1/3=1/(2(1+1)
II P(k) (P(k+1)
P(k+1): 1/(1(3)+1/(3(5)+...+ 1/[(2n-1)(2n+1)]+1/[(2n+1)(2n+3)]=(n+1)/(2n+3)
n/(2n+1)+1/[(2n+1)(2n+3)]=(n+1)/(2n+3)
[n(2n+3)+1]/[(2n+1)(2n+3)]=(n+1)/(2n+3)
(2n2+3n+1)/[(2n+1)(2n+3)]= (n+1)/(2n+3)
[(n+1)(2n+1)]/[(2n+3)(2n+1)]=(n+1)/(2n+3) (A)
I+II(P(n) (A) (n(1
Exemple de inegalitati rezolvate prin inductie
ex 1: p(n): 2n>n2 ; n(5
I P(5): 25>52 ( 32>25
II P(n) ( P(n+1): 2n+1>(n+1)2(
2n>n2 ((2 (2n+1 > 2n2 > (n+1)2
2n(2>2n2 ( 2n+1>2n2 (
Avem de dem ca 2n2>(n+1)2
( 2n2>n2+2n+1 ( n2-2n>1 (+1
(n2-2n+1>2 ( (n-1)2>2 (n(5
n(5 ( n-1(4 ( (n-1)2(16>2 (A)
I+II(P(n) adevarata (n(5
ex 2: p(n): 1/(n+2)+1/(n+2)+...+1/(2n)>13/24 n(2
I P(2): 1/3 +1/(2+2)>13/24 (7/12>13/24 ( 14/24 >13/24 (A)
II P(n+1): 1/(n+2)+1/(n+3)+..+1/(2n)+1/(2n+1)+1/(2n+2)>13/24
>13/24-1/(n+1)+1/(2n+1)+1/(2n+2)
=13/24-1/(2n+2)+1/(2n+1)
= 13/24 +1/[2(n+1)(2n+1)] (
1/[2(n+1)(2n+1)]>0 (13/24 +1/[2(n+1)(2n+1)]>13/24
I +II(P(n) adevarata (n(2
ex 3: p(n): 1/(n+1)+1/(n+2)+...+1/(3n+1)>1 n(1 ( 2n+1 termeni
I P(1): 1/(1+1)+1/(1+2)+1/(1+3)>1 ( 13/12>1 (A)
II P(n+1)=1/(n+2)+1/(n+3)+...+1/(3n+1)+1/(3n+2)+1/(3n+3)+1/(3n+4)
>1-1/(n+1)+1/(3n+2)+1/(3n+3)+1/(3n+4)
= 1+ ceva/[3(n+1)(3n+2)(3n+4)] >1(
ceva>0 ; 3(n+1)(3n+2)(3n+4)>0 (( (A)
I+II( P(n) adevarata (n(1
Alte exemple rezolvate prin inductie
ex 1: p(n): 10n+18n-28 : 27 n(0
I P(0): -27 :27 (A)
II P(n) (A) (P(n+1): 10n+1+18(n+1)-28 :27
10n+18n-28=27p(10n+1+18n-10=27q
10n= 27p -18n+28
10n+1+18n-10=10(10n+18n-10=10(27p -18n+28)+18n-10
=10(27p-180n+280+18n-10
=10(27p-16...


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