Quandaries and Queries


I'm a substitute teacher and I have had to give out the following problem for homework to a grade 12 class.  I have found the problem very difficult and being a non-math specialist, I would like you to give me the solution.
Find the nth term of the following sequence
4, 10, 28, 82
Many thanks


Hi Amanda,

You might notice that the terms are powers of 3 plus one and make an induction hypothesis that the nth term is 1+3n and proceed by induction.

If you don't spot this approach then you should look at differences but it's harder. The differences are, respectively, 6, 18 and 54 so one way to go is to take the next difference as 3(54) = 162 and so on, they're twice powers of 3. In this manner the kth difference is

3((k-i)st difference) = 32((k-2) difference) = ... = 3(k-1)(1st difference) = 2(3(k-1)).

Thus the nth term is

4 + (6+18+54+ ...) = 4 + 6(1+3+32+33+...+3(n-2)).

Now we have to sum a geometric series. We get a total of

4+6((3(n-1)-1)/2 = 4+3n-3 = 1+3n.



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