SOLUTION: I take any two numbers which I call a and b (lets say 2 and 3). These are my first 2 numbers. I then add them together to give my 3rd number (in this case 5), then to get my 4th nu

Question 535101: I take any two numbers which I call a and b (lets say 2 and 3). These are my first 2 numbers. I then add them together to give my 3rd number (in this case 5), then to get my 4th number I add the second and third numbers (8). My 5th number is the sum of the third and forth numbers (5 + 8 = 13). My 6th is the sum of the forth and fifth etc until I have 10 numbers (if I start with 2 and 3 they would be 2, 3, 5, 8, 13, 21, 34, 55, 89, 144) Add these 10 numbers together.I can find the answer in my head before you can using a calculator using algebra. Please help me work out this question i have to find out how to use algebra to find the answer.Please help. thank you.
Found 3 solutions by bucky, octogirl, Alan3354:
Answer by bucky(2189)   (Show Source): You can put this solution on YOUR website!
Your problem is about to disappear from the listing because it's becoming too old.
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It's a very interesting problem. I've done some work on it and I'm passing it along so you can think about it.
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Let's start by assuming, as you did, that you are given two numbers, A and B.
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Given that, let's start writing the terms in the series:
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#1. A
#2. B
#3. A + B
#4. (A + B) + B which is A + 2B ... [Note this is terms #3 plus #2]
#5. (A + 2B) + (A + B) which is 2A + 3B ... [terms #4 plus #3]
#6. (2A + 3B) + (A + 2B) which is 3A + 5B ... [terms #5 plus #4]
#7. (3A + 5B) + (2A + 3B) which is 5A + 8B ... [terms #6 plus #5]
#8. (5A + 8B) + (3A + 5B) which is 8A + 13B ... [terms #7 plus plus #6]
#9. (8A + 13B) + (5A + 8B) which is 13A + 21B ... [terms #8 plus #7)
10. (13A + 21B) + (8A + 13B) which is 21A + 34B ... [terms #9 plus #8]
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We could continue doing this for as many more terms as you want to, but let's stop here to develop an idea.
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Let's count the number of A's in each of these first ten terms:
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In term #1 there's 1, in term #2 there's 0, in term #3 there's 1, in term #4 there's 1, in term number 5 there's 2, in term #6 there's 3, in term #7 there's 5, in term # 8 there's 8, in term #9 there's 13, and finally in term #10 there's 21. Writing these numbers out like a series, we have for the first 10 terms the numbers of A's are: 1, 0, 1, 1, 2, 3, 5, 8, 13, 21. In a bit we'll do the same thing for the number of B's in the first ten terms. But first:
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The series 1, 1, 2, 3, 5, 8, 13, 21 ... is a very well known series (among math people) called the Fibonacci series. (Look it up in Wikipedia for further info.) Notice that it follows the same general pattern that you developed in your example which was 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, etc. You were only missing the first 1, 1 from having a complete Fibonacci series. The complete Fibonacci series can be totaled up relatively easily. You decide what term you want to stop at. Then figure out what the next two terms beyond that would be. Take that second term subtract 1 from it, and that's the total for the terms up to the term where you chose to stop. As an example. Suppose we had the Fibonacci series: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 and we wanted to total it up to and including the term 55. We see that the next two terms are 89 and 144. We take the second of these (the 144) and subtract 1 to get 143, and (check this out) that is the sum of all ten terms starting with the very first 1 and continuing through 55.
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Now let's go back to the series for the number of A's we had. We had 1 plus zero plus the first 8 terms of the Fibonacci series. Since it only involves 8 terms of the Fibonacci series we can just add those on a calculator to get an answer of 1+1+2+3+5+8+13+21 = 54. So the total number of A's we have in the first 10 terms is 1 + 0 + 54 = 55 A's.
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Now let's do the same kind of analysis for the number of B's in our answer. If we count the number of B's in each of the 10 terms we listed above we get the following series: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34. On a calculator this totals to 88. So for 10 terms we have a total of 88 B's.
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Notice that we did not specify what A and B were. They could be any two numbers and we would have for the first 10 terms 55 A's and 88 B's. So the short way to find the total of the first 10 terms would be to first find 55 times A, then find 88 times B, and finally add those two products.
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Let's use it on the example you gave in explaining the problem. You selected A = 2 and B = 3. This leads to 55 times 2 = 110 and 88 times 3 = 264 for a total of 374. If you will add up the 10 terms you gave in your example, you will find that the total of 2, 3, 5, 8, 13, 21, 34, 55, 89, and 144 does equal 374. Works for me (I think).
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This method [(55 times A) + (88 times B)] will work as long as you are looking at 10 terms. If you increase the number of terms, especially if it's a big increase) you can do the same type of analysis. The answer will change somewhat. The multiplier of A will become 1 + 0 + the sum of the Fibonacci series up to two less terms than the total number of terms you specify as the end. [Notice that for our work we specified and used 10 terms and we got 1 + 0 + the sum of the Fibonacci series out to 8 Fibonacci terms.] The number of B terms will be the sum of the number of Fibonacci terms out to one less than the total number of terms we specified. [In our analysis above we had specified 10 terms and can show that the sum of the number of B's is the same as the sum of the first 9 Fibonacci terms: 1+1+2+3+5+8+13+21+34 = 88].
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So if you want to specify more than 10 terms, for the number of A's you need to total out the Fibonacci series to the two less Fibonacci terms than the specified number of total terms and then add 1. And for the number of B's just total the Fibonacci series out to 1 less term than the specified number of total terms. Multiply the first sum by A and the second sum by B.
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Hope this isn't all too confusing to you. But it's food for thought, and as long as you limit yourself to a 10 term series it should be fairly straight-forward. If you increase the term series, you get a little deeper into the Fibonacci series than you may want to.
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Good luck. Hopefully this will help, and check my work. Thanks.
Answer by octogirl(7)   (Show Source): You can put this solution on YOUR website!
you change all the numbers into equation like 2=a 3=b... then you add all a to get a total and add all the b. you'll get an answer of 55a+88b there is a relation between this answer and the 7th term which is divide by eleven. try it with numbers it works eg. for 2 and 3 as a and b, 34 is the 7th term if you multiply that by eleven you get the total which is 374
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Look up the Fibonacci Series.
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