Use complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$












2












$begingroup$


Define the sequence of integers $a_0, a_1, a_2, cdots$ as follows



$$ a_i =
begin{cases}
i+1 & 0 leq i leq 2 \
a_{i-1} + a_{i-2} + 2a_{i-3} & i > 2 \
end{cases}
$$



Use complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$



I will prove $a_n < 2^n, forall n geq 2$ by using complete induction



Base Case: Three cases $n = 2, 3, 4$



let $n = 2$



$a_{n} = a_{2} = 2 + 1 = 3 < 4 = 2^2 = 2^n$ By definition, and holds



let $n = 3$



$a_{n} = a_{3} = 3 + 2 + 1 = 6 < 8 = 2^3 = 2^n$ By definition, and holds



let $n = 4$



$a_{n} = a_{4} = 4 + 3 + 2 = 9 < 16 = 2^4 = 2^n$ By definition, and holds



Inductive step: let $n > 4$. Suppose $a_j < 2^j$ whenever $2 leq j < n$. [Inductive hypothesis]



What to prove: $a_n < 2^n$



$a_{n} = a_{i-1} + a_{i-2} + 2a_{i-3}$ [By definition]



$< 2^{n-1} + 2^{n-2} + 2^{n-3+1}$ [By Inductive hypothesis]



$= 2^{n-1} + 2^{n-2} + 2^{n-2}$



$= 2^{n-1} + 2^{n-2 + 1}$



$= 2^{n-1 + 1}$



$= 2^n$



as wanted.



Would this be correct?






induction







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Define the sequence of integers $a_0, a_1, a_2, cdots$ as follows



    $$ a_i =
    begin{cases}
    i+1 & 0 leq i leq 2 \
    a_{i-1} + a_{i-2} + 2a_{i-3} & i > 2 \
    end{cases}
    $$



    Use complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$



    I will prove $a_n < 2^n, forall n geq 2$ by using complete induction



    Base Case: Three cases $n = 2, 3, 4$



    let $n = 2$



    $a_{n} = a_{2} = 2 + 1 = 3 < 4 = 2^2 = 2^n$ By definition, and holds



    let $n = 3$



    $a_{n} = a_{3} = 3 + 2 + 1 = 6 < 8 = 2^3 = 2^n$ By definition, and holds



    let $n = 4$



    $a_{n} = a_{4} = 4 + 3 + 2 = 9 < 16 = 2^4 = 2^n$ By definition, and holds



    Inductive step: let $n > 4$. Suppose $a_j < 2^j$ whenever $2 leq j < n$. [Inductive hypothesis]



    What to prove: $a_n < 2^n$



    $a_{n} = a_{i-1} + a_{i-2} + 2a_{i-3}$ [By definition]



    $< 2^{n-1} + 2^{n-2} + 2^{n-3+1}$ [By Inductive hypothesis]



    $= 2^{n-1} + 2^{n-2} + 2^{n-2}$



    $= 2^{n-1} + 2^{n-2 + 1}$



    $= 2^{n-1 + 1}$



    $= 2^n$



    as wanted.



    Would this be correct?






    induction







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Define the sequence of integers $a_0, a_1, a_2, cdots$ as follows



      $$ a_i =
      begin{cases}
      i+1 & 0 leq i leq 2 \
      a_{i-1} + a_{i-2} + 2a_{i-3} & i > 2 \
      end{cases}
      $$



      Use complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$



      I will prove $a_n < 2^n, forall n geq 2$ by using complete induction



      Base Case: Three cases $n = 2, 3, 4$



      let $n = 2$



      $a_{n} = a_{2} = 2 + 1 = 3 < 4 = 2^2 = 2^n$ By definition, and holds



      let $n = 3$



      $a_{n} = a_{3} = 3 + 2 + 1 = 6 < 8 = 2^3 = 2^n$ By definition, and holds



      let $n = 4$



      $a_{n} = a_{4} = 4 + 3 + 2 = 9 < 16 = 2^4 = 2^n$ By definition, and holds



      Inductive step: let $n > 4$. Suppose $a_j < 2^j$ whenever $2 leq j < n$. [Inductive hypothesis]



      What to prove: $a_n < 2^n$



      $a_{n} = a_{i-1} + a_{i-2} + 2a_{i-3}$ [By definition]



      $< 2^{n-1} + 2^{n-2} + 2^{n-3+1}$ [By Inductive hypothesis]



      $= 2^{n-1} + 2^{n-2} + 2^{n-2}$



      $= 2^{n-1} + 2^{n-2 + 1}$



      $= 2^{n-1 + 1}$



      $= 2^n$



      as wanted.



      Would this be correct?






      induction







      share|cite|improve this question











      $endgroup$




      Define the sequence of integers $a_0, a_1, a_2, cdots$ as follows



      $$ a_i =
      begin{cases}
      i+1 & 0 leq i leq 2 \
      a_{i-1} + a_{i-2} + 2a_{i-3} & i > 2 \
      end{cases}
      $$



      Use complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$



      I will prove $a_n < 2^n, forall n geq 2$ by using complete induction



      Base Case: Three cases $n = 2, 3, 4$



      let $n = 2$



      $a_{n} = a_{2} = 2 + 1 = 3 < 4 = 2^2 = 2^n$ By definition, and holds



      let $n = 3$



      $a_{n} = a_{3} = 3 + 2 + 1 = 6 < 8 = 2^3 = 2^n$ By definition, and holds



      let $n = 4$



      $a_{n} = a_{4} = 4 + 3 + 2 = 9 < 16 = 2^4 = 2^n$ By definition, and holds



      Inductive step: let $n > 4$. Suppose $a_j < 2^j$ whenever $2 leq j < n$. [Inductive hypothesis]



      What to prove: $a_n < 2^n$



      $a_{n} = a_{i-1} + a_{i-2} + 2a_{i-3}$ [By definition]



      $< 2^{n-1} + 2^{n-2} + 2^{n-3+1}$ [By Inductive hypothesis]



      $= 2^{n-1} + 2^{n-2} + 2^{n-2}$



      $= 2^{n-1} + 2^{n-2 + 1}$



      $= 2^{n-1 + 1}$



      $= 2^n$



      as wanted.



      Would this be correct?






      induction




      induction






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 8:45







      Tree Garen

















      asked Dec 1 '18 at 7:53









      Tree GarenTree Garen

      36219




      36219






















          1 Answer
          1






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          0












          $begingroup$

          The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But it doesn't hold for $n = 0$ or $n = 1$?
            $endgroup$
            – Tree Garen
            Dec 1 '18 at 8:01












          • $begingroup$
            You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
            $endgroup$
            – José Carlos Santos
            Dec 1 '18 at 8:06












          • $begingroup$
            I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
            $endgroup$
            – Tinler
            Dec 1 '18 at 8:14












          • $begingroup$
            Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
            $endgroup$
            – José Carlos Santos
            Dec 1 '18 at 8:26











          Your Answer





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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          0












          $begingroup$

          The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But it doesn't hold for $n = 0$ or $n = 1$?
            $endgroup$
            – Tree Garen
            Dec 1 '18 at 8:01












          • $begingroup$
            You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
            $endgroup$
            – José Carlos Santos
            Dec 1 '18 at 8:06












          • $begingroup$
            I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
            $endgroup$
            – Tinler
            Dec 1 '18 at 8:14












          • $begingroup$
            Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
            $endgroup$
            – José Carlos Santos
            Dec 1 '18 at 8:26
















          0












          $begingroup$

          The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But it doesn't hold for $n = 0$ or $n = 1$?
            $endgroup$
            – Tree Garen
            Dec 1 '18 at 8:01












          • $begingroup$
            You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
            $endgroup$
            – José Carlos Santos
            Dec 1 '18 at 8:06












          • $begingroup$
            I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
            $endgroup$
            – Tinler
            Dec 1 '18 at 8:14












          • $begingroup$
            Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
            $endgroup$
            – José Carlos Santos
            Dec 1 '18 at 8:26














          0












          0








          0





          $begingroup$

          The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.






          share|cite|improve this answer









          $endgroup$



          The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 7:59









          José Carlos SantosJosé Carlos Santos

          161k22127232




          161k22127232












          • $begingroup$
            But it doesn't hold for $n = 0$ or $n = 1$?
            $endgroup$
            – Tree Garen
            Dec 1 '18 at 8:01












          • $begingroup$
            You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
            $endgroup$
            – José Carlos Santos
            Dec 1 '18 at 8:06












          • $begingroup$
            I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
            $endgroup$
            – Tinler
            Dec 1 '18 at 8:14












          • $begingroup$
            Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
            $endgroup$
            – José Carlos Santos
            Dec 1 '18 at 8:26


















          • $begingroup$
            But it doesn't hold for $n = 0$ or $n = 1$?
            $endgroup$
            – Tree Garen
            Dec 1 '18 at 8:01












          • $begingroup$
            You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
            $endgroup$
            – José Carlos Santos
            Dec 1 '18 at 8:06












          • $begingroup$
            I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
            $endgroup$
            – Tinler
            Dec 1 '18 at 8:14












          • $begingroup$
            Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
            $endgroup$
            – José Carlos Santos
            Dec 1 '18 at 8:26
















          $begingroup$
          But it doesn't hold for $n = 0$ or $n = 1$?
          $endgroup$
          – Tree Garen
          Dec 1 '18 at 8:01






          $begingroup$
          But it doesn't hold for $n = 0$ or $n = 1$?
          $endgroup$
          – Tree Garen
          Dec 1 '18 at 8:01














          $begingroup$
          You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
          $endgroup$
          – José Carlos Santos
          Dec 1 '18 at 8:06






          $begingroup$
          You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
          $endgroup$
          – José Carlos Santos
          Dec 1 '18 at 8:06














          $begingroup$
          I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
          $endgroup$
          – Tinler
          Dec 1 '18 at 8:14






          $begingroup$
          I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
          $endgroup$
          – Tinler
          Dec 1 '18 at 8:14














          $begingroup$
          Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
          $endgroup$
          – José Carlos Santos
          Dec 1 '18 at 8:26




          $begingroup$
          Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
          $endgroup$
          – José Carlos Santos
          Dec 1 '18 at 8:26


















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