Posts: 167
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Joined: Jun 2010
21 May 12, 10:00AM
(This post was last modified: 21 May 12, 02:12PM by tempest.)
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 :
![[Image: png.latex?S%20=%20F_%7B3%7D%20+%20F_%7B6%7D%20+%...+%20F_%7B3n%7D]](http://latex.codecogs.com/png.latex?S%20=%20F_%7B3%7D%20+%20F_%7B6%7D%20+%20...%20+%20F_%7B3n%7D)
By the definition of fibonacci numbers, we know that
![[Image: png.latex?F_%7B3k%7D%20=%20F_%7B3k-1%7D%20+%20F_%7B3k-2%7D]](http://latex.codecogs.com/png.latex?F_%7B3k%7D%20=%20F_%7B3k-1%7D%20+%20F_%7B3k-2%7D)
which means
![[Image: png.latex?S%20=%20F_%7B1%7D%20+%20F_%7B2%7D%20+%...20F_%7B3n-1%7D]](http://latex.codecogs.com/png.latex?S%20=%20F_%7B1%7D%20+%20F_%7B2%7D%20+%20F_%7B4%7D%20+%20F_%7B5%7D%20+%20...%20+%20F_%7B3n-2%7D%20+%20F_%7B3n-1%7D)
now adding both sides together we get
![[Image: png.latex?2S%20=%20\sum_%7Bi=1%7D%5E%7B3n%7D%20F_%7Bi%7D]](http://latex.codecogs.com/png.latex?2S%20=%20\sum_%7Bi=1%7D%5E%7B3n%7D%20F_%7Bi%7D)
Now one can easily prove this identity with induction over
![[Image: png.latex?n]](http://latex.codecogs.com/png.latex?n)
:
![[Image: png.latex?\sum_%7Bi=1%7D%5E%7Bn%7D%20F_%7Bi%7D%20=%20F_%7Bn+2%7D-1]](http://latex.codecogs.com/png.latex?\sum_%7Bi=1%7D%5E%7Bn%7D%20F_%7Bi%7D%20=%20F_%7Bn+2%7D-1)
Using that we get
![[Image: png.latex?2S%20=%20F_%7B3n+2%7D-1%20=%3E%20S...n+2%7D-1%7D%7B2%7D]](http://latex.codecogs.com/png.latex?2S%20=%20F_%7B3n+2%7D-1%20=%3E%20S%20=%20\frac%7BF_%7B3n+2%7D-1%7D%7B2%7D)
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.