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August 19th, 2016, 08:37 AM   #1
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Even Fibonacci Numbers

Hello!

I don't really have good knowledge about Series in mathematics and I can't
solve the problem below:

Find the sum of the even terms of Fibonacci's function which are less
than four million, with first two terms 1 and 2
.

We also can have the help of a computer.

So I thought that this would be easy if every time I calculate the next term
and if that term divided by 2 gives me a modulus of 0 (which means that
it's an even number), then I will add it to my sum.

But this is impossible because I need to do this four million times - 1
even for a computer it takes a lot time to calculate that!!!!

So I was wondering if using mathematical series, can I find the sum
easily? How can I succeed then?

Thanks

Last edited by skipjack; August 19th, 2016 at 03:13 PM.
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August 19th, 2016, 08:44 AM   #2
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Actually, I'm wrong; if I do this four million times - 1, I will take terms that are
a lot greater than four million because every new term in Fibonacci becomes a lot greater than the previous term. So I don't really know.

Please help!!!

Last edited by skipjack; August 19th, 2016 at 02:58 PM.
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August 19th, 2016, 10:52 AM   #3
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A quick program in Mathematica, but it's written in standard style so it can be easily translated to the language of your choice.

fprev = 1;
fcurr = 2;
sum = 0;

While[fcurr < 4000000,
tmp = fcurr;
fcurr = fcurr + fprev;
fprev = tmp;
If[EvenQ[fcurr], sum += fcurr];
];

sum (* this just displays the final sum *)
4613730
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Last edited by romsek; August 19th, 2016 at 10:57 AM.
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August 19th, 2016, 03:30 PM   #4
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Quote:
Originally Posted by babaliaris View Post
Find the sum of the even terms of Fibonacci's function which are less than four million, . . . with first two terms 1 and 2.
This specification is unclear. Why does it state that the first term is 1, given that 1 isn't even? If you are being asked to sum the values 2, 8, 34, etc., where each value is less than four million, you have only 11 values to sum, so you wouldn't need a computer . . . you could do that by hand.
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August 20th, 2016, 05:32 AM   #5
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Quote:
Originally Posted by romsek View Post
A quick program in Mathematica, but it's written in standard style so it can be easily translated to the language of your choice.

fprev = 1;
fcurr = 2;
sum = 0;

While[fcurr < 4000000,
tmp = fcurr;
fcurr = fcurr + fprev;
fprev = tmp;
If[EvenQ[fcurr], sum += fcurr];
];

sum (* this just displays the final sum *)
4613730
I wrote the exact same code, but in my computer the loop could never finish.
So I thought maybe I could find the sum using series.

Actually I did one mistake :P . I didn't put while (fcurr < 4000000) but instead I had a counter (counter < 4000000, counter++)
and this is that drove me saying 4 million times - 1 which was a wrong decision. I will try that and tell you the results.

Ok I did it and it worked in an instant. But I must correct that you forgot to sum the 2. So the result is 4613732.

I also would like to know, is this possible without a computer using mathematic techniques
(and I'm thinking of series of course :P )? Or we must wait for someone to invent new mathematical techniques

Last edited by skipjack; August 21st, 2016 at 01:17 AM.
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August 20th, 2016, 05:42 AM   #6
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Quote:
Originally Posted by skipjack View Post
This specification is unclear. Why does it state that the first term is 1, given that 1 isn't even? If you are being asked to sum the values 2, 8, 34, etc., where each value is less than four million, you have only 11 values to sum, so you wouldn't need a computer . . . you could do that by hand.
It gives you the 1 because you can't calculate the next term without it, even if it's not an even number. Then you have to choose only the even terms that occur and sum them.

Maybe you didn't understand quite well.

The 11th term in Fibonacci is the number 144 which is not the last term that is less than 4 million. I must sum all the even terms up to the last which is less than 4 million.

Last edited by skipjack; August 20th, 2016 at 06:13 AM.
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August 20th, 2016, 06:18 AM   #7
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The 11th even term is 3524578, which is the last even term under 4 million.

The even terms are 2, 8, 2 + 4*8 = 34, 8 + 4*34 = 144, etc.
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August 20th, 2016, 06:26 AM   #8
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Quote:
Originally Posted by skipjack View Post
The 11th even term is 3524578, which is the last even term under 4 million.

The even terms are 2, 8, 2 + 4*8 = 34, 8 + 4*34 = 144, etc.
You meant the 11th even term my bad
Then there is actually a way to figure this out without the help of
a computer!!! Thank you very much! Maths is really awesome!

I wish I could be better in maths but it's so difficult to gain this knowledge.

I'm trying 2 years to pass calculus in the university that I'm studying,
but I failed the exams 3 times because of Series and maybe integrals too. To tell the truth, I wasn't also much of a reader.

Last edited by skipjack; August 20th, 2016 at 07:49 AM.
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August 20th, 2016, 08:00 AM   #9
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This is an algebra question, so I've moved it to the Algebra section of the site.
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August 20th, 2016, 08:42 AM   #10
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Quote:
Originally Posted by skipjack View Post
This is an algebra question, so I've moved it to the Algebra section of the site.
Ok no problem, didn't know that
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