The Fibonacci sequence follows the simple rule of adding the previous two numbers to get the next.

0,1,1,2,3,5,8,13,21,34, etc

That is stated in your problem as D_n = D_(n-1) + D_(n-2) (for example, 21=13+8)

Now, trying to decipher the problem, I believe it is saying that when calculating the formula 1 + 4w(1 + w + n), where n follows a fibonacci sequence, the difference between each successive answer also follows the fibonacci sequence.

Let's plug in some real numbers to get an idea of this.

Let us try w=3:

n    T_n   Difference
0      49    
1      61    12
1      61     0
2      73    12
3      85    12
5    109    24
8    145    36
13   205    60
21   301    96
34   457   156
 
So, we have n following the Fibonacci sequence, we have calculated 1 + 4w(1 + w + n) for w=3 and different values of n, and we have calculated the difference. Interestingly, the difference also follows the rule that the previous two numbers add up to make the next. (12+0=12) (12+12=24) etc


If you believe I have stated the problem correctly, please let me know, and we can then work on the proof.

well....

OK, I believe we can look at the problem like this: we want to know the difference between two successive n-values for any w (this difference will be the "D" value mentioned in your questiom)

First let's expand out the function: 1+4w(1+w+n) = 1+4w+4w^2+4wn

Let us see what the difference between two n-values is (call them n_1 and n_2)

   1+4w+4w^2+4wn_1
- 1+4w+4w^2+4wn_2
= 0+  0+       0+4w(n_2-n_1)

= 4w(n_2-n_1)   

So, that would be the solution for one difference "D" (remember the D_n = D_(n-1) + D_(n-2) formula?)

Let us try to "plug this into" that the D_n = D_(n-1) + D_(n-2) formula. For ease of writing this on the screen I will use actual numbers, but you can substitute (n-1), etc:

4w(n_3-n_2) = 4w(n_2-n_1) + 4w(n_1-n_0)

simplifying:
n_3-n_2 = (n_2-n_1) + (n_1-n_0)
n_3-n_2 = n_2-n_0

And that last formula is indeed a Fibonacci sequence which we can prove by playing with it:

substitute n_2 for n_2=n_1+n_0

n_3-n_2 = (n_1+n_0)-n_0

Simplifying:

n_3-n_2 = n_1

or
n_3 = n_2 + n_1

So, I have "shown" how this can be proved. You just need to neaten it up a little before calling it a proof, OK?