Induction equation proof











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Lets define $a_0=a_1=1$ and every $k$ what is bigger than $1$ and is integer,
$$a_k=a_{k-1}+2a_{k-2}.$$ Prove with induction, that every integer that is $kgeq0$
the equation



$$a_k=dfrac{(2^{k+1}+(-1)^k)}{3}.$$






induction







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    – jayant98
    Nov 22 at 21:25















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Lets define $a_0=a_1=1$ and every $k$ what is bigger than $1$ and is integer,
$$a_k=a_{k-1}+2a_{k-2}.$$ Prove with induction, that every integer that is $kgeq0$
the equation



$$a_k=dfrac{(2^{k+1}+(-1)^k)}{3}.$$






induction







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  • Show your work too!
    – jayant98
    Nov 22 at 21:25













up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











Lets define $a_0=a_1=1$ and every $k$ what is bigger than $1$ and is integer,
$$a_k=a_{k-1}+2a_{k-2}.$$ Prove with induction, that every integer that is $kgeq0$
the equation



$$a_k=dfrac{(2^{k+1}+(-1)^k)}{3}.$$






induction







share|cite|improve this question















Lets define $a_0=a_1=1$ and every $k$ what is bigger than $1$ and is integer,
$$a_k=a_{k-1}+2a_{k-2}.$$ Prove with induction, that every integer that is $kgeq0$
the equation



$$a_k=dfrac{(2^{k+1}+(-1)^k)}{3}.$$






induction




induction






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edited Nov 22 at 21:45









Yadati Kiran

1,327418




1,327418










asked Nov 22 at 21:22









Joe574

1




1












  • Show your work too!
    – jayant98
    Nov 22 at 21:25


















  • Show your work too!
    – jayant98
    Nov 22 at 21:25
















Show your work too!
– jayant98
Nov 22 at 21:25




Show your work too!
– jayant98
Nov 22 at 21:25










3 Answers
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Try the following theorem:




If $x_1$ and $x_2$ are solutions of $x^2-pt-q=0$, then $a_k=c_1x_1^k+c_2x_2^k$ is solution of the recurrence $a_k-pa_{n-1}-qa_{k-2}=0$.




After this, you can try $k=0,1$ in the geral solution to obtain a particluar answer, your answer.






share|cite|improve this answer




























    up vote
    1
    down vote













    The characteristics equation helps us here which is: $$x^2=x+2$$and has two roots $$x_1=2\x_2=-1$$therefore$$a_n=a2^n+b(-1)^n$$ by substituting the initial conditions we obtain what we want






    share|cite|improve this answer




























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      For $n=0$ the base case we have $a_0=dfrac{(2^{0+1}+(-1)^0)}{3}=1$. Let us assume the result is true for $n=k$ i.e. $a_k=dfrac{(2^{k+1}+(-1)^k)}{3}$. We shall prove for $n=k+1$. $a_{k+1}=a_k+2a_{k-1}=dfrac{(2^{k+1}+(-1)^k+2cdot(2^k+(-1)^{k-1}))}{3}=dfrac{(2^{k+2}+(-1)^{k-1}(-1+2))}{3}=dfrac{(2^{k+2}+(-1)^{k-1}(-1)^2)}{3}=dfrac{(2^{k+2}+(-1)^{k+1})}{3}$.



      Hence by principle of mathematical induction the result is true for all $n$.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

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        active

        oldest

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        up vote
        1
        down vote













        Try the following theorem:




        If $x_1$ and $x_2$ are solutions of $x^2-pt-q=0$, then $a_k=c_1x_1^k+c_2x_2^k$ is solution of the recurrence $a_k-pa_{n-1}-qa_{k-2}=0$.




        After this, you can try $k=0,1$ in the geral solution to obtain a particluar answer, your answer.






        share|cite|improve this answer

























          up vote
          1
          down vote













          Try the following theorem:




          If $x_1$ and $x_2$ are solutions of $x^2-pt-q=0$, then $a_k=c_1x_1^k+c_2x_2^k$ is solution of the recurrence $a_k-pa_{n-1}-qa_{k-2}=0$.




          After this, you can try $k=0,1$ in the geral solution to obtain a particluar answer, your answer.






          share|cite|improve this answer























            up vote
            1
            down vote










            up vote
            1
            down vote









            Try the following theorem:




            If $x_1$ and $x_2$ are solutions of $x^2-pt-q=0$, then $a_k=c_1x_1^k+c_2x_2^k$ is solution of the recurrence $a_k-pa_{n-1}-qa_{k-2}=0$.




            After this, you can try $k=0,1$ in the geral solution to obtain a particluar answer, your answer.






            share|cite|improve this answer












            Try the following theorem:




            If $x_1$ and $x_2$ are solutions of $x^2-pt-q=0$, then $a_k=c_1x_1^k+c_2x_2^k$ is solution of the recurrence $a_k-pa_{n-1}-qa_{k-2}=0$.




            After this, you can try $k=0,1$ in the geral solution to obtain a particluar answer, your answer.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 22 at 21:33









            DiegoMath

            2,0491021




            2,0491021






















                up vote
                1
                down vote













                The characteristics equation helps us here which is: $$x^2=x+2$$and has two roots $$x_1=2\x_2=-1$$therefore$$a_n=a2^n+b(-1)^n$$ by substituting the initial conditions we obtain what we want






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  The characteristics equation helps us here which is: $$x^2=x+2$$and has two roots $$x_1=2\x_2=-1$$therefore$$a_n=a2^n+b(-1)^n$$ by substituting the initial conditions we obtain what we want






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    The characteristics equation helps us here which is: $$x^2=x+2$$and has two roots $$x_1=2\x_2=-1$$therefore$$a_n=a2^n+b(-1)^n$$ by substituting the initial conditions we obtain what we want






                    share|cite|improve this answer












                    The characteristics equation helps us here which is: $$x^2=x+2$$and has two roots $$x_1=2\x_2=-1$$therefore$$a_n=a2^n+b(-1)^n$$ by substituting the initial conditions we obtain what we want







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 22 at 21:38









                    Mostafa Ayaz

                    13.5k3836




                    13.5k3836






















                        up vote
                        0
                        down vote













                        For $n=0$ the base case we have $a_0=dfrac{(2^{0+1}+(-1)^0)}{3}=1$. Let us assume the result is true for $n=k$ i.e. $a_k=dfrac{(2^{k+1}+(-1)^k)}{3}$. We shall prove for $n=k+1$. $a_{k+1}=a_k+2a_{k-1}=dfrac{(2^{k+1}+(-1)^k+2cdot(2^k+(-1)^{k-1}))}{3}=dfrac{(2^{k+2}+(-1)^{k-1}(-1+2))}{3}=dfrac{(2^{k+2}+(-1)^{k-1}(-1)^2)}{3}=dfrac{(2^{k+2}+(-1)^{k+1})}{3}$.



                        Hence by principle of mathematical induction the result is true for all $n$.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          For $n=0$ the base case we have $a_0=dfrac{(2^{0+1}+(-1)^0)}{3}=1$. Let us assume the result is true for $n=k$ i.e. $a_k=dfrac{(2^{k+1}+(-1)^k)}{3}$. We shall prove for $n=k+1$. $a_{k+1}=a_k+2a_{k-1}=dfrac{(2^{k+1}+(-1)^k+2cdot(2^k+(-1)^{k-1}))}{3}=dfrac{(2^{k+2}+(-1)^{k-1}(-1+2))}{3}=dfrac{(2^{k+2}+(-1)^{k-1}(-1)^2)}{3}=dfrac{(2^{k+2}+(-1)^{k+1})}{3}$.



                          Hence by principle of mathematical induction the result is true for all $n$.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            For $n=0$ the base case we have $a_0=dfrac{(2^{0+1}+(-1)^0)}{3}=1$. Let us assume the result is true for $n=k$ i.e. $a_k=dfrac{(2^{k+1}+(-1)^k)}{3}$. We shall prove for $n=k+1$. $a_{k+1}=a_k+2a_{k-1}=dfrac{(2^{k+1}+(-1)^k+2cdot(2^k+(-1)^{k-1}))}{3}=dfrac{(2^{k+2}+(-1)^{k-1}(-1+2))}{3}=dfrac{(2^{k+2}+(-1)^{k-1}(-1)^2)}{3}=dfrac{(2^{k+2}+(-1)^{k+1})}{3}$.



                            Hence by principle of mathematical induction the result is true for all $n$.






                            share|cite|improve this answer












                            For $n=0$ the base case we have $a_0=dfrac{(2^{0+1}+(-1)^0)}{3}=1$. Let us assume the result is true for $n=k$ i.e. $a_k=dfrac{(2^{k+1}+(-1)^k)}{3}$. We shall prove for $n=k+1$. $a_{k+1}=a_k+2a_{k-1}=dfrac{(2^{k+1}+(-1)^k+2cdot(2^k+(-1)^{k-1}))}{3}=dfrac{(2^{k+2}+(-1)^{k-1}(-1+2))}{3}=dfrac{(2^{k+2}+(-1)^{k-1}(-1)^2)}{3}=dfrac{(2^{k+2}+(-1)^{k+1})}{3}$.



                            Hence by principle of mathematical induction the result is true for all $n$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 22 at 21:32









                            Yadati Kiran

                            1,327418




                            1,327418






























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