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Fibonacci sums (Posted on 2004-07-15) Difficulty: 3 of 5
The Fibonacci series 0, 1, 1, 2, 3, 5, 8, 13, in which each number is the sum of the two previous, is defined as F(0)=0, F(1)=1, and F(n)=F(n-1)+F(n-2) for n>1.

What is the sum of F(0)+F(1)+F(2)+...+F(k)?
What is the sum of F(0)^2+F(1)^2+F(2)^2+...+F(k)^2?

See The Solution Submitted by Federico Kereki    
Rating: 3.7143 (7 votes)

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Solution Solution To Part B Comment 13 of 13 |
(In reply to Solution To Part A by K Sengupta)

Solution To Part B

We observe that:
F(1) = F(3) - 1 = 1 = 1*1 = F(1)*F(2)
F(1)^2 + F(2)^2 = 1+1 = 2 = 1*2 = F(2)*F(3)
F(1)^2 + F(2)^2 + F(3)^2= 1+1+ 4 = 6 = F(3)*F(4)......(ii)

This leads us to conjecture that:

Sum(i = 1 To k) = F(k)*F(k+1).......(**)

Let the above conjecture be true for k = p

Then,

Sum (i = 1 To p)[F(i)^2] = F(p)*F(p+1)
Or, F(1)^2 + F(2)^2 + F(3)^3+ ......+ F(p)^2 + F(p+1)^2
= F(p+1) [F(p) + F(p+1)]
= F(p+1)*F(p+2)
Or, Sum (i = 1 To p+1)[F(i)^2] = F(p+1)*F(p+2)

Thus the conjecture is true for k = p+1. Since, in terms of (ii),
the conjecture holds for k = 1, 2, 3; it now follows that the
relationship Sum(i = 1 To k) = F(k)*F(k+1) holds for any positive integer value of k.


  Posted by K Sengupta on 2007-05-31 06:14:01
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