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

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Theorem

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

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


Then:

$\ds \forall n \in \N_{>0}: \sum_{j \mathop = 1}^n 2^{j - 1} L_j = 2^n F_{n + 1} - 1$

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 $\forall n \in \N_{>0}$, let $\map P n$ be the proposition:

$\ds \sum_{j \mathop = 1}^n 2^{j - 1} L_j = 2^n F_{n + 1} - 1$


Basis for the Induction

$\map P 1$ 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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\ds \sum_{j \mathop = 1}^k 2^{j - 1} L_j = 2^k F_{k + 1} - 1$


Then we need to show:

$\ds \sum_{j \mathop = 1}^{k + 1} 2^{j - 1} L_j = 2^{k + 1} F_{k + 2} - 1$


Induction Step

This is our induction step:

\(\ds \sum_{j \mathop = 1}^{k + 1} 2^{j - 1} L_j\) \(=\) \(\ds \sum_{j \mathop = 1}^k 2^{j - 1} L_j + 2^k L_{k + 1}\)
\(\ds \) \(=\) \(\ds 2^k F_{k + 1} - 1 + 2^k L_{k + 1}\) Induction hypothesis
\(\ds \) \(=\) \(\ds 2^k \paren {F_{k + 1} + L_{k + 1} } - 1\)
\(\ds \) \(=\) \(\ds 2^k \paren {F_{k + 1} + F_k + F_{k + 2} } - 1\) Lucas Number as Sum of Fibonacci Numbers
\(\ds \) \(=\) \(\ds 2^k \paren {F_{k + 2} + F_{k + 2} } - 1\)
\(\ds \) \(=\) \(\ds 2^{k + 1} F_{k + 2} - 1\)

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


Therefore:

$\ds \forall n \in \N_{>0}: \sum_{j \mathop = 1}^n 2^{j - 1} L_j = 2^n F_{n + 1} - 1$

$\blacksquare$


Sources

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