proof that $n^2 + 2n \leq 5^n $

Question

How do I prove the following statement using induction? $(n \in \mathbb{N})$

$$n^2 + 2n \leq 5^n $$

So for $n = 1$ it is true, since $1 + 2 \leq 5$.

So now:

Assumption: $ n^2 + 2n \leq 5^n $

Hypothesis (inductive step): $(n+1)^2 + 2(n+1) \leq 5^{n+1} $

Also I know that: $$5^n < 5^n + 2n$$

Then I get: $$ n^2 < 5^n$$

So: $$\frac{n^2}{5^n} < 1$$

And I have to prove that (hypothesis): $$ \frac{(n+1)^2}{5^{(n+1)}} < 1 $$ $$ \frac{n^2 + 2n + 1}{5 \cdot 5^n} < 1 $$

And since $$ \frac{n^2}{5 \cdot 5^n} < \frac{n^2 + 2n + 1}{5 \cdot 5^n}$$

So $$ \frac{n^2}{5 \cdot 5^n} < 1 $$

From assumption I have : $$\frac{n^2}{5^n} < 1$$

so hypothesis is also true : $$\frac{n^2}{5^n} < 5$$

Did I make it right?

Answer 1

This can be a better solution: $$(n+1)^2+2(n+1) \le 5^n + 1 + 2n + 2 = 5^n+2n+3 \le 5^n+5^n+3\cdot 5^n=5\cdot 5^n=5^{n+1}$$

Answer 2

A simpler proof is this.

$(n+1)^2+2(n+1)=n^2+2n+1+2n+2=(n^2+2n)+(2n+3)\leq 5^n+5^n=2\cdot 5^n<5^{n+1}$

Answer 3

First prove this enequality (this is much easier than first problem): $$2n+3\le 4×5^n$$ now plus 2n+3 to both side of your assumption you get: $$n^2+2n+2n+3\le 5^n+2n+3 \\ \Rightarrow (n+1)^2+2(n+1)\le 5^n+4×5^n=5×5^n=5^{n+1}$$