Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Guess a closed form for the sum F 1 + F 3 + . . . + F 2n1 , where n > 0 is a nat

ID: 3664543 • Letter: G

Question

Guess a closed form for the sum F1 + F3 + . . . + F2n1 , where n > 0 is a natural number and (Fk )k0 are Fibonacci numbers. This closed form may involve Fibonacci numbers. Show by induction that your guess is correct.

Hint: Let us consider the values of these sums for small n’s for a clue. Here is a reminder of the initial segment of Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, . . . where the numbers Fi with odd indices i are in bold. Let us compute the smallest sums:

F1 = 1

F1 + F3 = 1 + 2 = 3

F1 + F3 + F5 = 3 + 5 = 8

F1 + F3 + F5 + F7 = 8 + 13 = 21

F1 + F3 + F5 + F7 + F9 = 21 + 34 = 55

The outcomes we obtain are all Fibonacci. And there is a pattern.

Explanation / Answer

From the small sums, we cud see a simple pattern.

F1 = 1 = F2

F1 + F3 = 1 + 2 =3 = F4

F1 + F3 + F5 = 3 + 5 = 8 = F6

F1 + F3 + F5 + F7 = 8 + 13 = F8

F1 + F3 + F5 + F7 + F9 = 21 + 34 = 55 = F10

Hence , we can guess the closed form for F1 + F3 + . . . + F2n1 = F2n

--------------------------------------------------------------------------------------------------------------

Proof my Mathematical Induction :

To prove : F(n): F1 + F3 + . . . + F2n1 = F2n where n > 0 is a natural number and (Fk)k0 are Fibonacci numbers

1. Basis or base case

n = 1.
F(1) = 1 = F(2) , F(1) holds  true

2.

Assume F(k) holds (for some unspecified value of k). It must then be shown that F(k + 1) holds, that is

Assume F(k) : F1 + F3 + . . . + F2k1 = F2k holds true.

To prove  F(k+1) holds true is  F(k) holds true.

that is, F(k+1) : F1 + F3 + . . . + F2k1+F2k+1 = F2k+2

Using the induction hypothesis that F(k) holds, the left-hand side can be rewritten to:

  F1 + F3 + . . . + F2k1+F2k+1 = F2k+2

   F2k + F2k+1 = F2k+2

Using definition of fibinocci numbers , Fm + Fm+1 = Fm+2 , sum of consecutive terms gives the next term in the sequence.

Thus , F2k + F2k+1 = F2k+2 is true.

thereby showing that indeed F(k + 1) holds.

Since both the basis and the inductive step have been performed, by mathematical induction on n, the statement F(n) holds for all natural numbers n>1.