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Show for all $n \in \mathbb{N}$ $(n\geq1):$ $10^n \gt 6n^2+n$

My solution:

Base case: For $n=1$

$10^1 \gt 6 \cdot 1^2+1$

Inductive hypothesis:

$10^n \gt 6n^2+n \Rightarrow 10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$

Inductive step:

$10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$

$\Rightarrow$ $10^{n+1} \gt 6(n^2+2n+1)+n+1$

$\Rightarrow$ $10^{n+1} \gt 6n^2+12n+6+n+1$

$\Rightarrow$ $10^{n+1} \gt 6n^2+13n+7$

$\Rightarrow$ $10^n \cdot 10^1 \gt 6n^2+13n+7$

$\Rightarrow$ $(6n^2+n)\cdot10 \gt 6n^2+13n+7$

$\Rightarrow$ $60n^2+10n \gt 6n^2+13n+7$

I am stuck at this point. What techniques or tricks are there to solve the rest?

It is of interest to me, because I am currently practicing a lot of exercises related to convergences, inequalities and mathematical induction.

Any hints guiding me to the right direction I much appreciate.

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  • $\begingroup$ Note that $n \geqslant 1$, so the last inequality would be...? $\endgroup$
    – xbh
    Aug 29 '18 at 16:15
  • $\begingroup$ Easier one, prove that no. of digits in $10^n$ are way more than $6n^2+n$. For $n>3$ at least and show for $n=1,2$ $\endgroup$ Aug 29 '18 at 16:18
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For the inductive step we can proceed as follow

$$10^{n+1}=10\cdot 10^n \stackrel{Ind.Hyp.}>10\cdot(6n^2+n) \stackrel{?}>6\cdot(n+1)^2+(n+1)$$

and the latter requires

$$10\cdot(6n^2+n) \stackrel{?}>6\cdot(n+1)^2+(n+1)$$

$$60n^2+10n \stackrel{?}>6n^2+12n+6+n+1$$

$$54n^2-3n-7 \stackrel{?}>0$$

which can be shown to be true by quadratic formula or again by induction.

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  • $\begingroup$ Thanks, whats the name of the technique that you have used at the end? $\endgroup$
    – SAINT
    Aug 29 '18 at 16:27
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    $\begingroup$ @KeJie That way has not a particular name but I like that way to proceed starting from $10^{n+1}=10\cdot 10^n$ and then using the inductive hypotesis to obtain $10^{n+1}>10\cdot(6n^2+n)$ which of course (by hypotesis) is true. Finally we need that $10^{n+1}>10\cdot (6n^2+n)>6\cdot(n+1)^2+(n+1)$ and the inequality $(6n^2+n)>6\cdot(n+1)^2+(n+1)$ is the key point for the induction step. By algebraic step we find that it is equivalent to $54n^2-3n-7 >0$ which is true. $\endgroup$
    – user
    Aug 29 '18 at 16:31
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    $\begingroup$ @KeJie Your derivation wasn't completely wrong, indeed you finally obtain the same condition but your steps are mixed up and the derivation is not completely clear. We need to separate clearly what we are assuming true as hypotesis and what we need to prove. $\endgroup$
    – user
    Aug 29 '18 at 16:34
  • $\begingroup$ @KeJie- Since, $n^2>n$ and $n\ge1$. We can replace $n^2$ by $n$ and further $n$ by $1$ without changing the inequality $\endgroup$ Aug 29 '18 at 16:37
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You almost finished. Now you can transfer all to the left and group: $$60n^2+10n \gt 6n^2+13n+7 \iff \\ 54n^2-3n-7>0 \iff \\ 44n^2+3n^2-3n+7n^2-7>0 \iff \\ 44n^2+3n(n-1)+7(n^2-1)>0,$$ which is true because $n>1$.

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Hint: Multiplying $$10^n>6n^2+n$$ by $$10$$ we get

$$10^{n+1}>60n^2+10n$$ it remaines to prove that

$$60n^2+10n>6(n+1)^2+n+1$$ Can you do this? and this is $$54n^2-3n-7>0$$ which is clearly true for $$n\geq 1$$

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  • $\begingroup$ Thats the point where I don't know how to continue. $\endgroup$
    – SAINT
    Aug 29 '18 at 16:19
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$$60n^2+10n \gt 6n^2+13n+7 \iff 54n^2>3n+7$$

Note that $$ 54n^2 \ge 54n=3n+51n\ge 3n+51>3n+7$$

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  • $\begingroup$ Should it be $3n$ instead of $13n$? $\endgroup$
    – xbh
    Aug 29 '18 at 16:24
  • $\begingroup$ @xbh Thanks for the correction.You are right, I edited my answer. $\endgroup$ Aug 29 '18 at 18:07
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Left to show:

$54n^2-3n -7 >0$, $n \ge 1.$

$54n^2-3n-7 \ge 54 n^2 -3n^2-7=$

$51n^2 - 7 >0$.

Above inequality true for $n \ge 1$.

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