Why Can't the Sum of Any m Consecutive Fibonacci Numbers Always Be Odd?

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The discussion centers on the impossibility of finding a positive integer m such that the sum of every m consecutive Fibonacci numbers is odd. It is established that for m = 3, the sums of the sets of consecutive Fibonacci numbers, such as {1, 1, 2} and {1, 2, 3}, yield even results. The inquiry seeks a proof that this pattern holds for all m greater than 3, asserting that at least one subset of m consecutive Fibonacci numbers will always produce an even sum.

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Fibonacci numbers are the sequence 1,1,2,3,5,18,13,21... where after the initial two 1's, each number in the sequence is the sum of the previous two. Prove that there is no positive integer m such that the sum of every m consecutive Fibonacci numbers is odd.
Can anyone explain to me what's the underlined part mean? Thanks a lot for your help
 
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It means that if you pick a natural number, m, then taking all the sets of m consecutive fibonacci numbers will always guarantee that the sum of the elements of at least one of those subsets will be even.

For example, if m = 3 then the sets of consecutive fibonaccis would be

{1, 1, 2}, {1, 2, 3}, {2, 3, 5}, {3, 5, 8} and so forth. Respectively, the sums of the elements of these sets are 4, 6, , 10, 16 and so forth. Obviously, when m = 3 the sums will always be even. The question is will that always be the case when m > 3.
 
!The underlined part is asking for a proof that there does not exist a positive integer m such that the sum of every m consecutive Fibonacci numbers is odd. In other words, it is asking for evidence or reasoning to support the statement that it is impossible for there to be a pattern where the sum of any consecutive set of m Fibonacci numbers will always result in an odd number. This is a mathematical concept that can be proven using logical arguments and mathematical principles.
 

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