Question 389734
<pre><font size = 4 face = "batangche" color = "indigo"><b>

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   a<sub>n</sub>
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   a<sub>n</sub>
1   15  11
2   26  22 
3   48  44
4   92  88 
5  180

Observe that those differences are all multiples of 11

n   a<sub>n</sub>
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:

2<sup>0</sup>,2<sup>1</sup>,2<sup>2</sup>,2<sup>3</sup>

n   a<sub>n</sub>
1   15  11 = 1×11 = 2<sup>0</sup>×11 
2   26  22 = 2×11 = 2<sup>1</sup>×11
3   48  44 = 4×11 = 2<sup>2</sup>×11 
4   92  88 = 8×11 = 2<sup>3</sup>×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 a<sub>n+1</sub> from the
previous term a<sub>n</sub> we must add 2<sup>n-1</sup>×11.

So the recursion formula is

a<sub>1</sub> = 15, a<sub>n+1</sub> = a<sub>n</sub> + 2<sup>n-1</sup>×11

So to get the 6th term, a<sub>6</sub>, from the 5th term, a<sub>5</sub>,

we substitute 5 for n in a<sub>n+1</sub> = a<sub>n</sub> + 2<sup>n-1</sup>×11

a<sub>5+1</sub> = a<sub>5</sub> + 2<sup>5-1</sup>×11

a<sub>6</sub> = a<sub>5</sub> + 2<sup>4</sup>×11

Now we substitute 180 for a<sub>5</sub>

a<sub>6</sub> = 180 + 2<sup>4</sup>×11

a<sub>6</sub> = 180 + 16×11

a<sub>6</sub> = 180 + 176

a<sub>6</sub> = 356

Edwin</pre>