# Fibonacci Numbers

#### Bellarosa

1.Obtain the following formula for the alternating sum of Fibonacci Numbers

f(1)-f(2)+f(3)-f(4)+...+(-1)^(n+1) f(n)= (-1)^(n+1) [f(n-1) ]+1

2. Homework Equations

3. I subtracted the formulas for the sum of the first n Fibonacci numbers with odd indices together with the first n Fibonacci numbers with even indices

f(1)+f(3)+f(5)+...+f(2n-1) = f(2n)
- [f(2)+f(4)+f(6)+...+f(2n) = [f(2n+1) ]- 1]

= f(1)-f(2)+f(3)-f(4)+f(5)-f(6)+...+f(2n-1)-f(2n) = f(2n) - [f(2n+1)] +1

what I am having trouble with is obtaining the (-1)^(n+1) factor
I am thinking that if I factor the first two terms of the difference I would get the (-1) but as for its exponent I am not quite clear on how to derive it

Related Calculus and Beyond Homework Help News on Phys.org

### Want to reply to this thread?

"Fibonacci Numbers"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving