prove w/ full induction : $2^n + 4 > n^2 + 2n$. Let's skip the other steps for a moment, I'm just getting stuck with the proof the claim is: $2^{n+1} + 4 > (n+1)^2 + 2(n+1)$ and now the proof $$2^{n+1} + 4 > 2 * 2^n + 4$$ $$> 2 * (n^2 + 2n)$$ $$> 2n^2 + 4n$$ i am stuck here
Asked
Active
Viewed 72 times
1
3 Answers
1
It is true that for $n\geq5$
$$2^n+4>n^2+2n$$
Assume that for $n=k$ $$2^k+4>k^2+2k$$
Let's show for $n=k+1$
$$2^{k+1}+4>(k+1)^2+2(k+1)$$
Now
$$2(2^k+4)>2(k^2+2k)$$
$$2^{k+1}+4>2k^2+2k-4$$
$$2^{k+1}+4>2k^2+2k-4>(k+1)^2+2(k+1)$$
$$2k^2+2k-4>k^2+4k+3$$
$$k^2-2k-7>0$$
for $k\geq5$
$$(k-1)^2-8>0$$
medicu
- 4,482
Lion Heart
- 7,073
-
-
@mathslooser : $2(2^k+4)>2(k^2+2k)$, $2^{k+1}+8>2k^2+2k$, $2^{k+1}+4>2k^2+2k-4$ – Lion Heart Nov 13 '20 at 15:24
1
Please note this is not true for low values of $n \in \mathbb{Z+}$ up to $4$.
Just your induction step -
If $2^{n} + 4 > n^2 + 2n$ is true
we need to prove $\, 2^{n+1} + 4 > (n + 1)^2 + 2(n+1)$
$2^{n+1} + 4 = 2 \times 2^{n} + 8 - 4 = 2 (2^{n} + 4) - 4$
$ \gt 2n^2 + 4n - 4 = n^2 + (n+1)^2 + 2(n+1) - 7 = (n+1)^2 + 2(n+1) + (n^2 - 7)$
As $(n^2 - 7) \gt 0$ for $n \geq 3$, this is true.
Math Lover
- 51,819
0
The claim is $2^{n+1} + 4 > (n+1)^2 + 2(n+1)$
Assuming it is true for $n$, that is $2^n + 4 > n^2 + 2n\to 2^n>n^2+2n-4$ we have $$2^{n+1} + 4=2\cdot 2^n+4>2\left(n^2+2n-4\right)+4=2n^2+4n-4=(n^2+2n+1)+(2n+2)+n^2-6$$ $$>(n+1)^2+2(n+1);\;n>2$$ Proved
Raffaele
- 26,371
$signs around your mathematical symbols. For example,$2^n + 4 > n^2 + 2n$becomes $2^n + 4 > n^2 + 2n$. Please do experiment with it, as people will become less and less willing to format your questions for you. – user847970 Nov 13 '20 at 15:05