Answer to Question #202811 in Real Analysis for Innocent Sekgabi

Question

3ⁿ>2n²

Answer 1

We want to prove by induction that

$$3^n > 2n^2$$

Base case: n=1;

$$3^1 > 2(1)^2 \implies 3>2$$

Thus $S_1$ is true.

Inductive Step:

We assume it is true for n=k, so that

$$3^k > 2k^2$$

Inductive Step:

We will show that since it is true for n= k, then it is true for n=k+1.

$3^{k+1} = 3^k \cdot 3\\ \qquad > 2k^2 \cdot 3\\ \qquad > 6k^2\\ \qquad > 2k^2 + 4k^2\\ \qquad > 2[(k+1)^2-2k-1] + 4k^2\\ \qquad > 2(k+1)^2 -4k -2 + 4k^2\\ 3^{k+1} > 2(k+1)^2 +4k^2-4k-2\\$

Since

$$3^{k+1} > 2(k+1)^2 +4k^2-4k-2\\$$

then

$$3^{k+1} > 2(k+1)^2$$

which concludes the proof.