Sum of Sequence of Binomial Coefficients by Powers of 2/Proof 2

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\displaystyle \sum_{j \mathop = 0}^n 2^j \binom n j = 3^n$

$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

Basis for the Induction
$\map P 1$ is the case:
 * $\sum_{j \mathop = 0}^1 2^j \binom 1 j$

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_{j \mathop = 0}^k 2^j \binom k j = 3^k$

from which it is to be shown that:
 * $\displaystyle \sum_{j \mathop = 0}^{k + 1} 2^j \binom {k + 1} j = 3^{k + 1}$

Induction Step
This is the induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 0}: \displaystyle \sum_{j \mathop = 0}^n 2^j \binom n j = 3^n$