SOLUTION: 6 , 13 , 25 , 52 , 101 , ?

The method I show below will always work for any sequence.  However 
it may not be the answer your teacher wants.  If your teacher
wants a different answer, then he or she should tell you what
sort of answer you are to give, because sequence problems always 
have many possible different patterns for the nth terms which 
work for any finite number of given terms.  It is a bit much for 
a teacher to expect a student to try out thousands of possible 
nth terms all which work for the given terms in order to come up 
with the "simplest" one.  Teachers do not have to come up 
with the simplest one themselves, for they already have the answer 
from the source from which they got the sequence.  It is just not 
fair for teachers to ask students to do something that they 
themselves cannot do.  MY opinion!

That being said, the nth term which works for the given sequence,
I got by assuming an nth term of the form



and substituting n=1,2,3,4,5 and =6,13,25,52,12

I got:

, , , , 

Which gives:



Gettng a common denominator of 24, that becomes



To prove that this nth term works:

Substituting n = 1:









Substituting n = 2:









Substituting n = 3:









Substituting n = 4:









Substituting n = 5:









Substituting n = 6:









Substituting n = 7:









Substituting n = 8:









Substituting n = 9:









Substituting n = 10:









Substituting n = 11:









Substituting n = 12:









Substituting n = 13:









Substituting n = 14:









Substituting n = 15:









Substituting n = 16:









Substituting n = 17:









Substituting n = 18:









Substituting n = 19:









Substituting n = 20:










Edwin