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Simple proof of inequality by induction

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove [tex]2n+1< 2^n[/tex] for n=3,4,...

    2. Relevant equations

    3. The attempt at a solution
    P(3)= 6+1<2^3
    The base case holds, now assume true for [itex]P(k)[/itex]. Now consider [itex]P(k+1)[/itex]
    2(k+1) +1 = 2k+3 [/tex]
    2^{k+1} = 2^k \cdot 2

    So this is where I get stuck, I don't quite know how to finish the proof. As can be seen, the difference between the two sides of the inequality in P(k+1) is that the LHS has 2 added, while the RHS is multiplied by 2. Would it be sufficient to observe that in the base case, adding 2 to the LHS and multiplying the RHS by 2 still maintains the inequality and so necessarily it should hold for any larger n?
  2. jcsd
  3. Oct 12, 2009 #2


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    2(k+1) +1 = 2k+3 = (2k+1) + 1
    2^{k+1} = 2.2^k

    so starting with the case for n

    add 2 to both sides

    with one more step to show the case for n+1
  4. Oct 12, 2009 #3
    Well, it should be possible to show that [itex]2^k+2<2^k \cdot 2[/itex] for some k greater than whatever.

    [tex]2< 2\cdot 2^k - 2^k[/tex]
    [tex]2< (2-1)2^k[/tex]
    [tex]2<2^k[/tex]. From here take the log of both sides, and it shows that [tex]2^k +2 < 2^{k+1}[/tex] holds for all k > 1, which is fine because we're only considering from n=3. So since that's the case and [tex]2(n+1)+1 < 2^n +2 < 2^{n+1}[/tex] this should complete the proof?
  5. Oct 13, 2009 #4


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    i think that does it, but if you're happy upto

    then just using, that for n>1

    should pretty much complete above

    ps. i think itspretty much the same reasoning, though just a little clearer in order
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