Fibonacci and Lucas numbers related identities

Question

We know that $H_n$ = $L_n + mF_n$, where $n = 0$ or $n > 0$ is simply relation between Fibonacci sequence and generalized Fibonacci-Lucas sequence. Are there any methods to prove the following identities? I am new to this site and hopefully, I will have good replies.

1) Sum of the terms:

$$H_1 + H_2 + H_3 + \ldots H_n = H_{n+2} – (m+3)$$

2) For alternate terms, we see the following:

$$H_1 + H_3 + H_5 + \ldots +H_{2n-1} = H_{2n} – 2$$ and
$$H_2 + H_4 + H_6 + \ldots + H_{2n} = H_{2n+1} – (m + 1)\;.$$

3) At the same time, if we square each terms, we see the:

$$H_1^2 + H_2^2 + H_3^2 + \ldots + H_n^2 = H_n H_{n+1} – 2 (m + 1)\;.$$

4) Finally, this interesting identity is valid:

$$H_{2n} = \sum_{k=0}^{n}\binom{n}{k}H_{n-k}$$

Answer 1

1. By definition $$ \sum_{i=1}^n H_i = \sum_{i=1}^n L_i +m \sum_{i=1}^n F_i $$ where $L_{i\ge 1}=1,3,4,7,\ldots$ and $F_{i\ge 1} = 1,1,2,3,5,\ldots$. We have a well-known partial sum, see https://oeis.org/A027961 $$ \sum_{i=1}^n L_i = L_{n+2}-3 $$ and another see https://oeis.org/A000071 $$ \sum_{i=1}^n F_i = F_{n+2}-1 $$ Plugged into the first equation $$ \sum_{i=1}^n H_i = L_{n+2}-3 +m (F_{n+2}-1). $$ By definition $H_{n+2}=L_{n+2}+mF_{n+2}$, so $$ \sum_{i=1}^n H_i = H_{n+2}-mF_{n+2}- 3 +m (F_{n+2}-1) = H_{n+2}- 3 -m. $$ so the first sum of terms is proven.

2. For the sum of the odd-index terms by definition $$ \sum_{i=0}^{n-1} H_{2i+1} = \sum_{i=0}^{n-1} L_{2i+1} +m \sum_{i=0}^{n-1} F_{2i+1} $$ where with https://oeis.org/A004146 $$ \sum_{i=0}^{n-1} L_{2i+1} = L_{2n}-2 $$ and with https://oeis.org/A001906 $$ \sum_{i=0}^{n-1} F_{2i+1} = F_{2n}. $$ Therefore by definition $$ \sum_{i=0}^{n-1} H_{2i+1} =L_{2n}-2+m F_{2n} = H_{2n}-2, $$ so the second sum of term is also proven.

For the sum of the even-index terms by definition $$ \sum_{i=1}^{n} H_{2i} = \sum_{i=1}^{n} L_{2i} +m \sum_{i=1}^{n} F_{2i} $$ where with https://oeis.org/A188378 $$ \sum_{i=0}^{n} L_{2i} = L_{2n+1}+1. $$ Removing the inital $L_0=2$ yields $$ \sum_{i=1}^{n} L_{2i} = L_{2n+1}-1. $$ With https://oeis.org/A027941 $$ \sum_{i=1}^{n} F_{2i} = F_{2n+1}-1. $$ Therefore by definition $$ \sum_{i=1}^{n} H_{2i} =L_{2n+1}-1 +m(F_{2n+1}-1) =H_{2n+1}-1-m $$ so the third sum of terms is also proven.

3. For the sum of the squares by binomial expansion $$ \sum_{i=1}^n H_i^2 = \sum_{i=1}^n (L_i+mF_i)^2 = \sum_{i=1}^n L_i^2 +m^2 \sum_{i=1}^n F_i^2 +2m\sum_{i=1}^n L_iF_i. $$ Here by https://oeis.org/A005970 $$ \sum_{i=1}^n L_i^2 = L_n L_{n+1}-2. $$ By https://oeis.org/A001654 $$ \sum_{i=1}^n F_i^2 = F_n F_{n+1}. $$ and again by https://oeis.org/A027941 $$ \sum_{i=1}^n F_iL_i = F_{2n+1}-1. $$ Inserted above $$ \sum_{i=1}^n H_i^2 = L_n L_{n+1}-2 + m^2 F_n F_{n+1} +2m (F_{2n+1}-1). $$ Furthermore by definition $$ H_nH_{n+1} = (L_n+mF_n)(L_{n+1}+mF_{n+1}) = L_nL_{n+1} +mF_nL_{n+1} +mL_nF_{n+1}+m^2F_nF_{n+1}. $$ Subtracting the previous two equations yields $$ \sum_{i=1}^n H_i^2 - H_nH_{n+1} = L_n L_{n+1}-2 + m^2 F_n F_{n+1} +2m (F_{2n+1}-1) -[L_nL_{n+1} +mF_nL_{n+1} +mL_nF_{n+1}+m^2F_nF_{n+1}] $$ $$ = -2-2m +2m F_{2n+1} -mF_nL_{n+1} -mL_nF_{n+1} $$ By Bergot's formula in https://oeis.org/A001519 we know that the last 3 terms cancel, which establishes validity of 3).

The generating function of $H$ is $$ \sum_{i=0}^\infty H_{i} x^i = \sum_{i=0}^\infty L_{i} x^i +m \sum_{i=0}^\infty F_{i} x^i, $$ and the individual g.f. of the Lucas and Fibonacci sequenecs are very well known, see https://oeis.org/A000032 and https://oeis.org/A000045 $$ h(x)\equiv \sum_{i=0}^\infty H_{i} x^i = \frac{2-x}{1-x-x^2} +m\frac{x}{1-x-x^2}.\quad (1) $$

The generating function of the even-indexed $H$ is $$ \sum_{i=0}^\infty H_{2i} x^i = \sum_{i=0}^\infty L_{2i} x^i + m \sum_{i=0}^\infty F_{2i} x^i $$ where the individual g.f. are known by https://A005248 and https://oeis.org/A001906: $$ \sum_{i=0}^\infty L_{2i} x^i = \frac{2-3x}{1-3x+x^2}; $$ $$ \sum_{i=0}^\infty F_{2i} x^i = \frac{x}{1-3x+x^2}. $$ Therefore $$ h^{(2)}(x)\equiv \sum_{i=0}^\infty H_{2i} x^i = \frac{2-3x+mx}{1-3x+x^2}.\quad (2) $$ Furthermore the sequene of the inverse binomial transform of $H$ is $$ \sum_{k=0}^n\binom{n}{k}H_{n-k} = \sum_{k=0}^n\binom{n}{n-k}H_{n-k} = \sum_{k=0}^n\binom{n}{k}H_k.\quad (3) $$ It's a well-known property that the binomial transform replaces the argument of the generating function as follows see e.g. https://dx.doi.org/10.1016/0024-3795(94)00245-9 $$ b_n=\sum_{k=0}^n \binom{n}{k}a_k \leftrightarrow b(x) = \frac{1}{1-x}a(\frac{x}{1-x}). $$ If we apply this to (1) and (3) we have $$ \frac{1}{1-x}h(\frac{x}{1-x}) = \frac{2-3x+mx}{1-3x+x^2} $$ which matches exactly $h^{(2)}(x)$. Since the two generating functions match, the individual terms must also match, and 4) is also proven.