These problems are as easy as pie
1 - One can easily note that the first 2 fibonacci numbers are 1 and hence are odd. The third one is 2 which is even. The 4th and the 5th ones are odd too (odd+even = odd) and the 6th one is even. So the first n even numbers in fibonacci sequence have indices divisible by 3 (numbering them with natural numbers). Hence we need to calculate this : By the definition of fibonacci numbers, we know that which means now adding both sides together we get Now one can easily prove this identity with induction over :
Using that we get which is the desired answer.
2 - This can be calculated using Binomial coefficient which is represented in Pascal's triangle
Mod edit: use PNG rather than GIF.
1 - One can easily note that the first 2 fibonacci numbers are 1 and hence are odd. The third one is 2 which is even. The 4th and the 5th ones are odd too (odd+even = odd) and the 6th one is even. So the first n even numbers in fibonacci sequence have indices divisible by 3 (numbering them with natural numbers). Hence we need to calculate this : By the definition of fibonacci numbers, we know that which means now adding both sides together we get Now one can easily prove this identity with induction over :
Using that we get which is the desired answer.
2 - This can be calculated using Binomial coefficient which is represented in Pascal's triangle
Mod edit: use PNG rather than GIF.