Binomial coefficient summatory and Fibonacci numbers question

[olmoelisa]

There is a summatory of binomial coefficients which gives the Fibonacci

numbers:


(5 0) + (4 1) + (3 2) = 1 + 4 + 3 = 8 (Fib 7)

(9 0) + (8 1) + (7 2) + (6 3) + (5 4) = 1 + 8 + 21 + 20 + 5 = 55 (Fib 10)

If I alterne sum and subtraction I obtain 0, 1 or -1:

1 - 4 + 3 = 0
1 - 8 + 21 - 20 + 5 = -1

But what happen if I sum the 1st half and subtract the 2nd half of the

sequence?

That is:

1 + 4 -3 = 2 (or 1 - 4 -3 = - 6)
1 + 8 + 21 - 20 - 5 = 5 (or 1 + 8 - 21 - 20 - 5 = -37)

Any idea/paper/hint?

Thank you very much
Sandra

 

[Jhero]

There have been some studies that have looked at this type of operation. In particular, it is known as "Fibonacci-like identities" or "alternating sums of binomial coefficients". Some papers that have studied this include: 1. Yang, S.-T., & Tzeng, W.-Y. (2006). Alternating sums of binomial coefficients and Fibonacci numbers. The Fibonacci Quarterly, 44(2), 150–152. 2. Bussotti, A., & Tauraso, R. (2012). Alternating sums of binomial coefficients and Fibonacci numbers. Journal of Number Theory, 132(9), 1872–1885. 3. Lai, C.-C., & Yang, S.-T. (2008). Alternating sums of binomial coefficients and Fibonacci numbers II. The Fibonacci Quarterly, 46(1), 36–40. 4. Lai, C.-C., & Yang, S.-T. (2009). Alternating sums of binomial coefficients and Fibonacci numbers III. The Fibonacci Quarterly, 47(4), 317–324. 5. Lai, C.-C., & Yang, S.-T. (2012). Alternating sums of binomial coefficients and Fibonacci numbers IV. The Fibonacci Quarterly, 50(1), 1–7. I hope this helps!