Use mathematical induction to prove: 2n + 1 <= 2n, for n >= 3?

I have trouble proving this, for the three steps involved in induction p(3), p(n), p(n+1).

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  • 1 decade ago
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    OK assuming p(n) is true we are going to prove that p(n+1) true.

    P(n+1) says

    2(n+1)+1 <= 2(n+1)

    which is equivalent to

    2n+3<=2n+2

    which is equivalent to

    2n+1 <= 2n

    But this is p(n) and we already assume that p(n) is true. Therefore p(n+1) is also true.

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  • 4 years ago

    no need to use induction, so I'll show the easy, non-induction way first, then the way to do it by induction: (2n + 1)^2 = 4n^2 + 4n + 1, so ((2n + 1)^2)/4 = n^2 + n + 1/4 > n^2 + n >=n as desired. Now, by induction: [base case, n = 1]: 1 < [(3)^2]/4 = 9/4 [induction step] now, if n < ((2n+1)^2)/4, then n + 1 < ((2n+1)^2)/4 + 1 < ((2n+1)^2)/4 + 2n + 2 = ((2n+1)^2)/4 + (8n + 8)/4 = (4n^2 + 12n + 9)/4 = ((2n + 3)^2)/4 = ((2(n+1) + 1)^2)/4 so the inequality is also true for n + 1, as desired.

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