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Prove by mathematical induction that $n! ≥ n^3$ , for $n > 5$

The basic steps

  1. take number $6$

$$6!> 6 * 6 * 6$$

  1. assume $n=k$

$$k! \ge k^3$$

  1. For $n = k + 1 $

    $$(k+1)! \ge (k+1)^3$$

And I dont know how to continue

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    $\begingroup$ Please edit to include your efforts. As the problem suggests, try induction. $\endgroup$
    – lulu
    Commented Sep 6, 2022 at 18:38
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    $\begingroup$ Hint: $(k + 1)! = (k + 1) k!$ $\endgroup$
    – Dan
    Commented Sep 6, 2022 at 18:53
  • $\begingroup$ $6!=720>216=6^3$. Assume that $k!>k^3$. Then $(k+1)!=(k+1)k!>(k+1)k^3=k^4+k^3=(k+1)^3+(k^4-3k^2-3k-1)$. If we prove the $k^4-3k^2-3k+1\geq0$, for $k>5$ then we are done. Well, $k^4-3k^2-3k-1\geq 6k^3-3k^2-3k-1=3k^2(k-3)+3k(k^2-3)-1>-1$. This last part is quite relaxed. You can prove it in many ways. $\endgroup$
    – plop
    Commented Sep 6, 2022 at 18:55
  • $\begingroup$ Where did that (k+1)k3 come from wasnt it(k+1) to the power of 3? $\endgroup$
    – Math
    Commented Sep 6, 2022 at 19:21

2 Answers 2

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First prove that $k^4+k^3>(k+1)^3$: $$k^4+k^3>k^3+3k^2+3k+1$$ $$k^4-3k^2-3k-1>0$$ $$(k^4-1)-3k(k+1)>0$$ $$(k+1)(k^3-k^2+k-1)-3k(k+1)>0$$ $$(k+1)(k^3-k^2-2k-1)>0$$ $$(k+1)((k-1)^3 + 2k^2-5k)>0$$ This is clearly positive for $k>5$.

Now, $(k+1)!=(k+1)\cdot k! \ge (k+1)k^3=k^4+k^3>(k+1)^3$

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  • $\begingroup$ Where did that (k+1)k3 come from wasnt it(k+1) to the power of 3? – $\endgroup$
    – Math
    Commented Sep 6, 2022 at 19:22
  • $\begingroup$ We've assumed that $k! \ge k^3$ (step 2 of induction). $\endgroup$
    – eMathHelp
    Commented Sep 6, 2022 at 19:23
  • $\begingroup$ If $k! \ge k^3$, then (multiply both sides by $k+1$) $(k+1)k! \ge (k+1)k^3$. $\endgroup$
    – eMathHelp
    Commented Sep 6, 2022 at 19:29
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you can use this proof to justify why n! > n³ when n > 5...

for n > 5...
n*(n-1)(n-2)(n-3)(n-4) must be greater than nnn because numbers (n-3) and (n-4) are both greater than or equal to 2 and therefore increase the values of (n-1) and (n-2) by a factor of at least 2.
you can confirm that (n-1)
(n-4) > n and (n-2)*(n-3) > n when n > 5 by solving the inequity for n...

$(n-1)*(n-4) > n$
$n² - 5n + 4 > n$
$n² - 6n + 4 > 0$
$(n-3+\sqrt{5})(n-3-\sqrt{5}) > 0$
$n>3+\sqrt{5}$

you can also follow the same logic for $(n-2)*(n-3) > n$ and get $n>3+\sqrt{3}$ which is slightly higher than $n>3+\sqrt{5}$, but still lower than 6

now that you proved that $n*(n-1)*(n-2)*(n-3)*(n-4)>n^3$, you can use the knowledge that all $(n-5), (n-6), (n-7)...$ positive values that will be multiplied to $n*(n-1)*(n-2)*(n-3)*(n-4)$ will only make that value larger, to conclude that $n!>n^3$ when $n>5$

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