Question 1097251
<br>There is no way to "solve" this, or any other problem like this.<br>
You could put any number next and it would be a valid sequence.<br>
The only way to get the "right" answer is to ask the author of the problem what answer he had in mind.<br>
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 {{{p(n) = an^4+bn^3+cn^2+dn+e}}} that will produce the given sequence.<br>
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
{{{a+b+c+d+e=3}}}  <-- p(1)
{{{16a+8b+4c+2d+e = 7}}}  <-- p(2)
...
{{{625a+125b+25c+5d+e = 52}}}  <-- p(5)<br>
The calculations are straightforward, but very tedious.<br>
Or a graphing calculator with matrix capability can be used to make the work relatively easy.<br>
But the answer obtained using the degree 4 polynomial might not be the "right" one that the author of the problem wanted........<br><br>
I see that another tutor has already supplied an answer that is very similar to one I was going to add to my response....<br>
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<br>
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
{{{3*2^(n-1)}}}<br>
The values produced by that formula alone are
 3, 6, 12, 24, 48, ...<br>
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
{{{3*2^(n-1)+(n-1)}}}<br><br>
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.<br>
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.