Sum of Sequence of Product of Lucas Numbers with Powers of 2

Theorem
Let $L_k$ be the $k$th Lucas number.

Let $F_k$ be the $k$th Fibonacci number.

Then:
 * $\displaystyle \forall n \in \N, n \ge 1: \sum_{j=1}^n 2^{j-1} L_j = 2^n F_{n+1} - 1$

That is:
 * $2^0 L_1 + 2^1 L_2 + 2^2 L_3 + \cdots 2^{n-1} L^n = 2^n F_{n+1} - 1$

Proof
Proof by induction:

For all $\forall n \in \N, n \ge 1$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{j=1}^n 2^{j-1} L_j = 2^n F_{n+1} - 1$

Basis for the Induction

 * $P(1)$ is true, as this just says $L_1 = 1 = 2^1 \times F_1 - 1$ which follows directly from the definitions of Fibonacci numbers and Lucas numbers.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{j=1}^k 2^{j-1} L_j = 2^k F_{k+1} - 1$

Then we need to show:
 * $\displaystyle \sum_{j=1}^{k+1} 2^{j-1} L_j = 2^{k+1} F_{k+2} - 1$

Induction Step
This is our induction step:

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

Therefore:
 * $\displaystyle \forall n \ge 1: \sum_{j=1}^n 2^{j-1} L_j = 2^n F_{n+1} - 1$