Show that $10^n \gt 6n^2+n$

Question

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.

Answer 1

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.

Answer 2

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 $$n\geq 1$$

Answer 3

$$60n^2+10n \gt 6n^2+13n+7 \iff 54n^2>3n+7$$

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

Answer 4

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 $n>1$.

Answer 5

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 $n \ge 1$.