Question

Please help? Forces and sigma notation:

Answer 1

If it's too small, I'll rewrite it

Answer 2

\[F _{n}= F _{n - 1} + F _{n - 2}\]
\[F _{0} = 0 and F _{1} = 1\]
\[\sum_{-5}^{5}F _{n}\]

Answer 3

its still too small for my bad eyes :(

Answer 4

I rewrot it

Answer 5

@alexwee123 You have to find the sigma.

Answer 6

Someone please help!

Answer 7

I don't believe there is a quick and easy way to do this one... and since it's not many terms, I'd just try to find explicit values for \(F_{-5}\), \(F_{-4}\),...,\(F_{5}\) and add them up.

Answer 8

Alright thanks! I'll try it. If it helps, this is the answer they gave:
Fn = {…, -8, 5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89…}. ∑_(n=-5)^5▒F_n =16.
The answer was 16.

Answer 9

You can also try simplifying by progressive substitutions, not to say it's that much easier.\[\sum_{-5}^{5}F _{n}=F_{-5}+F_{-4}+F_{-3}+F_{-2}+F_{-1}+F_{0}+F_1+F_2+F_3+F_4+F_5\]Substituting \(F_5=F_4+F_3\) gives:\[\sum_{-5}^{5}F _{n}=F_{-5}+F_{-4}+F_{-3}+F_{-2}+F_{-1}+F_{0}+F_1+F_2+2F_3+2F_4\]Substituting \(F_4=F_3+F_2\) gives:\[\sum_{-5}^{5}F _{n}=F_{-5}+F_{-4}+F_{-3}+F_{-2}+F_{-1}+F_{0}+F_1+3F_2+4F_3\]Substituting \(F_3=F_2+F_1\) gives:\[\sum_{-5}^{5}F _{n}=F_{-5}+F_{-4}+F_{-3}+F_{-2}+F_{-1}+F_{0}+5F_1+7F_2\]Substituting \(F_2=F_1+F_0\) gives:\[\sum_{-5}^{5}F _{n}=F_{-5}+F_{-4}+F_{-3}+F_{-2}+F_{-1}+8F_{0}+12F_1\]Substituting \(F_{-3}=F_{-4}+F_{-5}\) and \(F_{0}=F_{-1}+F_{-2}\) gives:\[\sum_{-5}^{5}F _{n}=2F_{-3}+9F_{0}+12F_1\]I mean, now it's a little bit simpler, but you still have to calculate \(F_{-3}\) to evaluate. It looks like by the solution the implication is that you're just supposed to calculate the terms individually.

Answer 10

Thank you so much! At the beginning I had no idea how to tackle the problem. Thank you!