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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 questions? I am new to this site and hopefully, I will have good replies.

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

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

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Solutions to constant coefficients linear recurrence relations should be a linear combination of exponentials (with polynomial coefficients sometimes, but constant coefficients ideally), cf Binet's formula, which should allow these sorts of identities to be proven with eg geometric sum formula or the binomial theorem. –  anon Aug 17 '12 at 6:54
    
@Anon! great response. Thank you. I tried Binet's formula and there is some disturbance in producing the proof. I am looking to give one proof for all questions? is it possible? I mean deducing all the results from one to other. If not solve the third and fourth one. Thanks a lot for for a quick reply. –  vidyaojal Aug 17 '12 at 6:59
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Do you know the corresponding identities for the Fibonacci sequence and for the Lucas sequence? For example, do you know the identity for $F_1+F_2+\cdots+F_n$? If you know those, you can use them to get identities for your $H_n$. –  Gerry Myerson Aug 17 '12 at 7:22
    
@GerryMyerson! I got all those results by your hint. Thank U. –  vidyaojal Aug 17 '12 at 14:52
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"I got all those results by your hint." - then, please post an answer to your own question, and we can then check if your method is correct. –  J. M. Aug 17 '12 at 15:58

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