Metode de rationament

Metode de rationament

I.     636i83g ;     636i83g ;     636i83g ;     636i83g ;    Metoda directa( modus poneus) - pornind de la o propozitie A si folosind principiul silogismului demonstram ca o alta propozitie este adevarata.

II.     636i83g ;     636i83g ;     636i83g ;     636i83g ; Metoda indirecta – reducerea la absurd ce se bazeaza pe echivalenta  (p®q) º (ùq®ùp)

III.     636i83g ;     636i83g ;     636i83g ;   Metoda inductiei

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Metoda inductiei

A. Egalitati

P(n):  1+2+3++n=n(n+1)/2             n³1

I. P(1):  1=1 (A)

II. P(n)  (A)ÞP(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+IIÞ P(n) (A)    'n³1

Deci: 1+2+3+n = åk = n(n+1)/2 , n³1

P(n): 1² +2² +3² ++n²= n(n+1)(2n+1)/6       n³1

I. P(1):  1=1 (A)

II. P(n) (A) ÞP(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)² =( n+1)(n+2)(2n+3)/6

             (n+1)(n+2)(2n+3)/6=( n+1)(n+2)(2n+3)/6  (A)

I+IIÞP(n)  (A)         'n³1

12 +22 +32 + +n2 = åk2 = n(n+1)(2n+1)/6, n³1

1³+2³+3³++n³= [n(n+1)/2]²         n³1

I. P(1): 1=1 (A)

II. P(n) (A) ÞP(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+IIÞP(n)  (A)         'n³1

13+23+33++n3=åk3 = [n(n+1)/2]2, n³1

½a1+a2+..+an½£½a1½+½a2½++½an½   n³2

Exemple de egalitati rezolvate prin inductie

ex 1: S=1×4+2×7+3×10++n(3n+1)

             =åk(3k+1) = å3k²+k = å3k²+åk = 3åk²+å

             = 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  n³1

          I  P(1): 1C4=1(1+1)² Þ  4=4 (A)

          II  P(n) ÞP(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+IIÞP(n)  (A) 'n³1

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)] Þå1/[(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) ÞP(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+IIÞP(n) (A) 'n³1        

 Exemple de inegalitati rezolvate prin inductie

ex 1: p(n): 2ⁿ>n² ; n³5

         I P(5): 2⁵>5²   Þ 32>25

          II P(n) Þ P(n+1): 2ⁿ⁺¹>(n+1)²½

          2ⁿ>n²   ½×2                              ½2ⁿ⁺¹ > 2n² > (n+1)²

          2ⁿ×2>2n²   Þ          2ⁿ⁺¹>2n²    ½

          Avem de dem ca 2n²>(n+1)²

          Þ 2n²>n²+2n+1 Þ n²-2n>1 ½+1

          Þn²-2n+1>2     Þ (n-1)²>2  'n³5

          n³5 Þ n-1³4  Þ (n-1)²³16>2      (A)

          I+IIÞP(n) adevarata 'n³5

ex 2: p(n):   1/(n+2)+1/(n+2)++1/(2n)>13/24         n³2

          I P(2): 1/3 +1/(2+2)>13/24 Þ7/12>13/24 Þ 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)] ½

                             1/[2(n+1)(2n+1)]>0           ½13/24 +1/[2(n+1)(2n+1)]>13/24

           I +IIÞP(n) adevarata       'n³2

ex 3: p(n): 1/(n+1)+1/(n+2)++1/(3n+1)>1      n³1 Þ  2n+1 termeni

          I P(1): 1/(1+1)+1/(1+2)+1/(1+3)>1 Þ 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)]  >1½

                                 ceva>0 ; 3(n+1)(3n+2)(3n+4)>0 ½Þ (A)

            I+IIÞ P(n) adevarata     'n³1

Alte exemple rezolvate prin inductie

ex 1:   p(n): 10ⁿ+18n-28  : 27          n³0

           I P(0): -27 :27 (A)

           II P(n) (A) ÞP(n+1): 10ⁿ⁺¹+18(n+1)-28 :27

               10ⁿ+18n-28=27pÞ10ⁿ⁺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-162n+270=27(10p-6n+10)=27q

             I+IIÞP(n)  (A)            'n³0

ex 2:    daca n³10  atunci 2ⁿ>n³     n³1

            P(n): (1/2)×(3/4)×× (2n-1)/2n<1/(Ö2n+1)

            I P(1):1/2 <1/Ö3 Û  2>Ö3  (A)

            II P(n+1): 1/2×3/4×× (2n-1)/2n× (2n+1)/(2n+2)<1/(Ö2n+3)

                            <1/(Ö2n+1) × (2n+1)/(2n+2)< 1/(Ö2n+3)

             (Ö2n+1)/(2n+2)< 1/(Ö2n+3)

              Ö(2n+1)(2n+3)<2n+2

              4n²+8n+3<4n²+8n+4 Þ3<4     (A)

            I+IIÞP(n) (A) 'n³1

ex 3: P(n): 2ⁿ>n³    n³10

         I P(10): 2¹⁰>10³  Þ  1024>1000 (A)

         II P(n) (A) Þ P(n+1)

             P(n+1): 2ⁿ⁺¹>(n+1)³       ½

             2ⁿ>n³ ½×2  Þ2ⁿ⁺¹>2n³   ½2ⁿ⁺¹>2n³>(n+1)³

            Avem de dem 2n³>(n+1)³

            2n³-(n+1)³>0  Þ  (³Ö2)³n³-(n+1)³>0    Þ (n ³Ö2)³-(n+1)³>0

             Þ (n ³Ö2-n-1)[n ²Ö4+n ³Ö2(n+1)+(n+1)^2]>0

              n ²Ö4+n ³Ö2(n+1)+(n+1)² >0

              n ³Ö2-n-1 >0   ?

             Þ n(³Ö2-1)>1   n  ³10        ½

               ³Ö2>1,1   Þ   ³Ö2-1>0,1   ½ n(³Ö2-1)>1

       I+IIÞP(n)  (A) 'n³10