The other tutor found a GENERAL formula, not a RECURSIVE formula.

List those in a column beside the numbers n=1 through n=5

n   an
1   15
2   26 
3   48
4   92 
5  180

Out beside each number, write the difference between it
and the number just below it.

n   an
1   15  11
2   26  22 
3   48  44
4   92  88 
5  180

Observe that those differences are all multiples of 11

n   an
1   15  11 = 1×11 
2   26  22 = 2×11
3   48  44 = 4×11 
4   92  88 = 8×11
5  180

Next we observe that the numbers 1,2,4,8 are these powers of 2:

20,21,22,23

n   an
1   15  11 = 1×11 = 20×11 
2   26  22 = 2×11 = 21×11
3   48  44 = 4×11 = 22×11 
4   92  88 = 8×11 = 23×11
5  180

Finally we observe that the exponents of 2 are 1 less than the
values on n in the left-most column, and 1 less than n is n-1.

So we think this way :  To get the next term an+1 from the
previous term an we must add 2n-1×11.

So the recursion formula is

a1 = 15, an+1 = an + 2n-1×11

So to get the 6th term, a6, from the 5th term, a5,

we substitute 5 for n in an+1 = an + 2n-1×11

a5+1 = a5 + 25-1×11

a6 = a5 + 24×11

Now we substitute 180 for a5

a6 = 180 + 24×11

a6 = 180 + 16×11

a6 = 180 + 176

a6 = 356

Edwin