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What is proper way to prove $2^n + 3 \ge n^2$, for all integer $n$ greater than $0$ , using mathematical induction? This is what I had tried:

  • $n = 1; 1 + 3 \geq 1; 4 \geq 1$ OK
  • Precondition $n = k; 2^{(k + 1)} + 3 \geq (k + 1)^2$
    $2 \times 2 ^ k + 3 \geq k^2 + 2k + 1$
    but nothing comes out of it. Thank You in advance
Harsh Kumar
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Lukas Narus
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  • in my condition it is "for all integer", corrected question. Seen this answer, but could not use that solution – Lukas Narus Mar 06 '17 at 20:37
  • "for all integer" is false, for example $2^{-2} + 3 = \frac{1}{4} +3 \leq 4 = 2^2$ – Zain Patel Mar 06 '17 at 20:40
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    @Arnaldo well... you can use induction in two directions. If you can somehow prove that $P(n)\implies P(n+1)$ as well as prove that $P(n)\implies P(n-1)$ and show that there is at least one $n$ such that $P(n)$ is true you will have succesfully proven the statement $P$ for all integers. – JMoravitz Mar 06 '17 at 20:49
  • @Arnaldo Once use for this inductive technique is for intervals. You want to prove a property of a function holds on all of $\mathbb{R}$ and can prove that it holds on $[0,1]$. You then prove that if it holds on an interval it holds on both adjacent intervals and so by induction it holds across all of $\mathbb{R}$. This tends to be situations where you're using the derivative and looking at potential rates of growth of a function and can constrain its variation on the next interval over. – Stella Biderman Mar 06 '17 at 20:58
  • Sorry that wa my huge mistake. I wanted to say all positive integer (n > 0) – Lukas Narus Mar 06 '17 at 20:58

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