SOLUTION: Conjecture a formula for the nth term of {an} if the first ten terms of this sequence are as follows. a) 3, 11, 19,27,35,43,51,59,67, 75 c) 1,0,0, 1,0,0,0,0, 1,0 b) 5, 7, 11, 1

Question 1184407: Conjecture a formula for the nth term of {an} if the first ten terms of this sequence are as follows.
a) 3, 11, 19,27,35,43,51,59,67, 75 c) 1,0,0, 1,0,0,0,0, 1,0
b) 5, 7, 11, 19, 35,67, 131,259,515, 1027 d) 1, 3,4, 7, 11, 18,29,47, 76, 123
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

To start with, thank you for saying "conjecture" a formula for each sequence.

We can in fact only make conjectures, because any subsequent numbers in each sequence would form a valid sequence.

So in these problems you are being asked to do some simple calculations or logical reasoning to try to find what MIGHT be the pattern in each sequence.

Note in some of these sequences it will be possible to find patterns which allow us to find subsequent terms in the sequence; in some of them it will be possible to find a closed formula for the n-th term in the sequence; and in some of them both are possible.

a) 3, 11, 19,27,35,43,51,59,67, 75

This one is easy; there is a common difference of 8 between terms -- that is, each term is found from the preceding term via addition. And for this one the formula for the n-th term is easily obtained. Since the common difference is 8 and the first term is 3, the formula is t(n)=8n-5.

b) 5, 7, 11, 19, 35,67, 131,259,515, 1027

The difference between terms get larger later in the sequence, so the pattern probably has something to do with multiplication instead of addition.

Do some quick rough calculations and observe that each term is a bit less than 2 times the preceding term. So look for a pattern that finds each term by doubling the preceding term and then doing something else:

7=2(5)-3
11=2(7)-3
19=2(11)-3
...

The pattern seems to be "double and subtract 3".

With that pattern, it is easy to calculate subsequent terms.

However, I don't know how to find an explicit formula for the n-th term when the pattern is of that type.

c) 1,0,0, 1,0,0,0,0, 1,0

What I see here is 1 followed by two 0's, then another 1 followed by four 0's, then...

Continuing the pattern we might next have a 1 followed by six 0's; but it would also be a logical pattern for the sequence to continue with a 1 followed by eight 0's. That is, the pattern of the numbers of 0's following each 1 could be 2, 4, 6, 8, 10, ...; or it could be 2, 4, 8, 16, ....

Now if we look at the sequence differently, we can see what very well might be the pattern. The 1's are in positions 1, 4, and 9, with 0's elsewhere.

So my conjecture for this sequence has an easy formula for the n-th term: t(n)=1 if n is a perfect square; t(n)=0 otherwise.

d) 1, 3,4, 7, 11, 18,29,47, 76, 123

It should be easy to see that in this sequence each term is the sum of the two preceding terms, so it is like the Fibonacci sequence. Note there is only one Fibonacci sequence; its first two terms are 1 and 1. Any other sequence that starts with any other two first numbers and then uses the same pattern is a Lucas sequence.

For Lucas sequences it is easy to find subsequent terms in the sequence, but it is not easy to find a closed formula for the n-th term.

45 years ago I spent a summer at Portland State University studying Lucas sequences in detail in a graduate course, and I think I learned how to find closed form formulas for the n-th term of any Lucas sequence. But I have no recollection of any details....

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SOLUTION: Conjecture a formula for the nth term of {an} if the first ten terms of this sequence are as follows. a) 3, 11, 19,27,35,43,51,59,67­, 75 c) 1,0,0, 1,0,0,0,0, 1,0 b) 5, 7, 11, 1

SOLUTION: Conjecture a formula for the nth term of {an} if the first ten terms of this sequence are as follows. a) 3, 11, 19,27,35,43,51,59,67, 75 c) 1,0,0, 1,0,0,0,0, 1,0 b) 5, 7, 11, 1