Prove by induction that for all integers n greater than or equal to 2, 2^n+1 < 3^n
This is how I would do it:
For $\displaystyle n=2$, $\displaystyle 2^3 < 3^2$, $\displaystyle 8<9$, true for $\displaystyle n=2$
Assume for $\displaystyle n=k$, this gives you $\displaystyle 2^{k+1} < 3^k$
Now the induction step, prove for $\displaystyle n=k+1$
$\displaystyle 2^{k+2} < 3^{k+1}$
$\displaystyle 2 \cdot 2^{k+1} < 3 \cdot 3^{k}$
Dividing both sides by 2
$\displaystyle 2^{k+1} < \frac{3}{2} \cdot 3^{k}$
Therefore if $\displaystyle 2^{k+1} < 3^k$, then $\displaystyle 2^{k+1}$ is definitely $\displaystyle < \frac{3}{2} \cdot 3^{k}$, true for all $\displaystyle n=k+1$ if true for $\displaystyle n=k$
True for $\displaystyle n=2$, therefore true for all integers greater than or equal to 2.
QED