SOLUTION: solve this. 3 7 14 27 52 ......

Question 1097251: solve this.
3 7 14 27 52 ......
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

There is no way to "solve" this, or any other problem like this.

You could put any number next and it would be a valid sequence.

The only way to get the "right" answer is to ask the author of the problem what answer he had in mind.

It is always possible to find AN answer to a problem like this. Any sequence of n terms can be fitted by a polynomial of degree (n-1). So with your sequence of 5 numbers, there is a polynomial of the form that will produce the given sequence.

To find that polynomial, you simply form a system of 5 equations in the 5 unknowns a, b, c, d, and e. The 5 equations are
<-- p(1)
<-- p(2)
...
<-- p(5)

The calculations are straightforward, but very tedious.

Or a graphing calculator with matrix capability can be used to make the work relatively easy.

But the answer obtained using the degree 4 polynomial might not be the "right" one that the author of the problem wanted........

I see that another tutor has already supplied an answer that is very similar to one I was going to add to my response....

If you notice that the numbers are "approximately doubled" each time, you can figure out a pattern:
3*2 plus 1 gives you 7
7*2 plus 0 gives you 14
14*2 plus -1 gives you 27
27*2 plus -2 gives you 52

To get an explicit formula for the n-th term in the sequence, note that the first term is 3 and each term after that is approximately 2 times the previous term. Then the explicit formula is going to be something of the form

The values produced by that formula alone are
3, 6, 12, 24, 48, ...

Comparing those values to the actual values
3, 7, 14, 27, 52, ...
we see that an explicit formula for the n-th term in the sequence is

Finally, note that the sequence you get from the polynomial method I described first will be different from the one produced by this other method.

And while both sequences would be valid answers to the problem, it is far more likely that the intended answer is the one obtained by this second method.
Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!

Start with 3 Double 3, get 6, add +1, get 7. Double 7, get 14, add 0, get 14. Double 14, get 28, add -1 get 27. Double 27, get 54, add -2, get 52. Double 52, get 104, add -3, get 101. Double 101, get 202, add -4, get 198. Double 198, get 396, add -5, get 391. Double 391, get 782, add -6, get 776. Double 776, get 1552, add -7, get 1545. Double 1545, get 3090, add -8, get 3082. Double 3082, get 6164, add -9, get 6155. Double 6155, get 12310, add -10, get 12300. Double 12300, get 24600, add -11, get 24589. Double 24589, get 49178, add -12, get 49166. Double 49166, get 98332, add -13, get 98319. Double 98319, get 196638, add -14, get 196624. Double 196624, get 393248, add -15, get 393233. Double 393233, get 786466, add -16, get 786450. Double 786450, get 1572900, add -17, get 1572883. Double 1572883, get 3145766, add -18, get 3145748. Double 3145748, get 6291496, add -19, get 6291477. Double 6291477, get 12582954, add -20, get 12582934. Double 12582934, get 25165868, add -21, get 25165847. Here's that recursion formula: , Here's another recursion formula for the same sequence: , Did you want a general formula? Edwin

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