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

Proof
Proof by induction:

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

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 $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

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

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

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