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So, as I tried to solve this questions I used the following : $\newcommand{\aa}{\mathbf{a}}$ $\aa_1 =1,~~ \aa_2 = 1$ and we know that in Fibonacci sequence, $\aa_n = \aa_{n-1} +\aa_{n-2}. $ Hence, $\aa_3 = 2,~~ \aa_4 = 3,~~ \aa_5 = 5,~~ \aa_6 = 8$

We notice an Odd Odd Even pattern starting from $\aa_1$. Since we are assuming $100$ terms, we take $100/3$, which approximates to $33.333$ (do we round this off to 33?) Proceeding this way we know that there is one even in each of these $33$ terms. Thus, number of even terms = $33$.

So, odd terms $= 100 - 33 = 67$.

Is this the correct approach to this question?

Lutz Lehmann
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Jimjamlorde
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2 Answers2

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Odd, Odd, Even, Odd, Odd, Even, is a correct observation.

This is a valid approach.

Saketh Malyala
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    Do you want to add a sentence of why rounding is not ideal and how to consider the last term? – quarague Sep 17 '19 at 06:38
  • I am not sure, how do we even consider the last term? Reason I rounded was because I knew number of terms cannot be in decimals. – Jimjamlorde Sep 17 '19 at 07:17
  • @IshaanSingh Next time, when you have a more complex pattern, say Odd, Even, Odd, Odd, Even, Even lets say (length 6). first find the total number of repetitions in the first hundred terms (16x6) and then add on the next four (odd, even, odd, odd) – Saketh Malyala Sep 17 '19 at 15:34
  • so in your case, just take the first 33 (odd, odd, even) and then add the next odd (#100) for a total of 67 odd terms, or 33 even terms – Saketh Malyala Sep 17 '19 at 15:35
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You know that there are 33 full 3-cycles inside this 100 element segment. the 100th element starts a new cycle, and the cycle start is Odd, so that indeed you get 67 times Odd.

However, had you used the other convention $F_0=0$, $F_1=1$, etc. then the cycle would have been Even, Odd, Odd and the 100th element $F_{99}$ would be Even, giving $34$ times Even and $66$ times Odd. So simple rounding will not in all cases give the correct result.

Lutz Lehmann
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