SOLUTION: write a recursive formula for the sequence 15, 26, 48, 92, 180,... Then find the next term

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Question 389734: write a recursive formula for the sequence 15, 26, 48, 92, 180,... Then find the next term
Found 2 solutions by richard1234, Edwin McCravy:
Answer by richard1234(7193) About Me  (Show Source):
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Note that the difference between two consecutive terms a%5Bk%5D and a%5Bk%2B1%5D is an increasing geometric sequence. In general, we have the sequence:

a%5B1%5D+=+k

a%5B2%5D+=+k+%2B+pr%5E0+=+a%5B1%5D+%2B+pr%5E0

a%5B3%5D+=+k+%2B+pr%5E0+%2B+pr%5E1+=+a%5B2%5D+%2B+pr%5E1

a%5B4%5D+=+k+%2B+pr%5E0+%2B+pr%5E1+%2B+pr%5E2+=+a%5B3%5D+%2B+pr%5E2, etc.

It is seen that a%5Bi%5D+=+a%5Bi-1%5D+%2B+pr%5E%28i-2%29 for all i+%3E=+2. We can substitute p = 11 and r = 2 to obtain ,

a%5Bi%5D+=+a%5Bi-1%5D+%2B+11%2A2%5E%28i-2%29 where a%5B1%5D+=+15

Since a%5B5%5D+=+180, a%5B6%5D+=+180+%2B+11%2A2%5E4+=+356

Note that a recursive sequence is defined on previous terms. It is possible to write each expression a%5Bi%5D in terms of i and other constants, but it wouldn't be recursive. Using the sum of a geometric sequence, we can get

a%5Bi%5D+=+15+%2B+11sum%282%5Ej%2C+j+=+0%2C+i-2%29+=+15+%2B+11%282%5E%28i-1%29+-+1%29

Checking, we have a%5B1%5D+=+15, a%5B2%5D+=+15+%2B+11+=+26, a%5B3%5D+=15+%2B+11%283%29+=+48,etc. which is the same sequence as the recursion.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!


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