Show that $S_L= L_n^2L_{n+1} -L_1L_2L_3-L_2L_3L_4-\cdots L_{n-2}L_{n-1}L_n$, $n\ge3$

Question

$2,1,3,4,7,11,...$ for $n=0,1,2,3,4,...$; It is the Lucas numbers

Let the sum of the cube of Lucas series be $S_L$

$n\ge3$

$S_L=2^3+1^3+3^3+4^3+\cdots+L_n^3$,

show that it has a form of $S_L=L_n^2L_{n+1}-L_1L_2L_3-L_2L_3L_4-\cdots-L_{n-2}L_{n-1}L_n$

I try: I can't think of any simple identities to use.

This is the only one might have some sort of link to it,

$L_{n+1}^3+L_{n+2}^3={L_{n+3}\over 2}(L_n^2+L_{n+1}^2+L_{n+2}^2)$

Any further hints?

Answer 1

In the given form, we can easily prove it via induction. Verify the case $n = 3$ by hand, and for $n \geqslant 4$, use the recurrence to get

\begin{align} L_n^2L_{n+1} &= L_n^2(L_n + L_{n-1}) \\ &= L_n^3 + L_nL_{n-1}L_n \\ &= L_n^3 + L_nL_{n-1}(L_{n-1} + L_{n-2}) \\ &= L_n^3 + L_{n-1}^2L_n + L_nL_{n-1}L_{n-2}, \end{align}

and thus with the induction hypothesis

\begin{align} \sum_{k = 0}^n L_k^3 &= \Biggl(\sum_{k = 0}^{n-1} L_k^3\Biggr) + L_n^3 \\ &= \Biggl(L_{n-1}^2 L_n - \sum_{m = 1}^{n-3} L_mL_{m+1}L_{m+2}\Biggr) + L_n^3 \tag{I.H.} \\ &= \Biggl(L_{n-1}^2 L_n - \sum_{m = 1}^{n-3} L_mL_{m+1}L_{m+2}\Biggr) + \bigl(L_n^2L_{n+1} - L_{n-1}^2L_n - L_{n-2}L_{n-1}L_n\bigr) \tag{see above} \\ &= L_n^2L_{n+1} - \sum_{m = 1}^{n-2} L_mL_{m+1}L_{m+2}. \end{align}