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Lots of people sent in the solution that the sixth term of the Fibonacci sequence starting with 2 and 38 is 196 and you found other sequences with 196 as one of the terms. Exactly how many other Fibonacci sequences contain the term 196? A lot of solutions as Jimmy rightly pointed out!
We are only looking for positive whole numbers. The terms increase quickly so 196 has to be one of the first few terms.The simplest Fibonacci sequence is:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 253, \ldots
and we denote the nth term of this sequence by F(n).
Starting with the terms a, b (for a and b positive whole
numbers and a < b) we get the general Fib sequence:
a, b, a+b, a+2b, 2a+3b, 3a+5b, 5a+8b, 8a+13b,
13a+21b, 21a+34b, 34a+55b, 55a+89b, 89a+144b,
\ldots
The nth term of the general Fib sequence f(n) = aF(n-2) + bF(n-1) and note that, if the term 196 occurs in the sequence, it can't be beyond the twelfth term as after that the terms are too large.
Here are some sequences containing 196.
Sequences with 196 as the first term
196, b, 196+b, 196+2b, \ldots | for b > 196 |
a, 196, \ldots | for 1 < a < 195 |
1, 195, 196, \ldots
2, 194, 196, \ldots
...
97, 99, 196, \ldots
98, 98, 196, \ldots etc
So far we see that there are infinitely many sequences with 196
as the first term; exactly 195 with 196 as the second term;
exactly 98 with 196 as the third term.
To find all the remaining sequences containing 196 we have to
find whole numbers a and b where:
a + 2b = 196
or
2a + 3b = 196
or
3a + 5b = 196
etc.
In general we have to find whole number values of a and b
satisfying
aF(n-2) + bF(n-1) = 196.
and so we need to find whole number solutions to these equations
for n = 4, 5, 6, \ldots12.
We shall consider one remaining case and leave the rest to the
reader.
For n = 6 we seek values of a and b such that 3a + 5b = 196
There are no solutions for a = 1 because then b would not be a
whole number. We have already seen that a = 2 and b = 38 gives
196 as the sixth term. For larger values of a we have to take
smaller values of b. For a = 3 or 4 or 5 or 6 there are
again no solutions because b has to be a whole number.
For a = 7 we have:
21 + 5b | = | 196 |
5b | = | 175 |
b | = | 35 |
giving the sequence 7, 35, 42, 77, 119, 196, \ldots
To find the remaining solutions for n = 6 we increase a by
steps of 5 and decrease b by steps of 3.
There are five solutions for n = 6, which are:
You can use the same method to find the solutions for n = 4, 5, 7, \ldots 12
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