Question 828588
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2 7 22 67 202

Each term is 1 more than 3 times the preceding term.

A recursive rule would be "to get the next term, multiply the previous term
by 3 then add 1": 

{{{a[1]=2}}}, {{{a[n+1]=3a[n]+1}}}

But let's see if we can get the general term:

We make the sequences of differences between successive 
terms to see if they follow a pattern

  2    7-2 =   5 = 5×3<sup>0</sup>
  7   22-7 =  15 = 5×3<sup>1</sup>
 22  67-22 =  45 = 5×3<sup>2</sup>
 67 202-67 = 135 = 5×3<sup>3</sup>
202

1st term = 2
2nd term = 2+5×3<sup>0</sup> = 7
3rd term = 2+5×3<sup>0</sup>+5×3<sup>1</sup> = 22
4th term = 2+5×3<sup>0</sup>+5×3<sup>1</sup>+5×3<sup>2</sup> = 67
5th term = 2+5×3<sup>0</sup>+5×3<sup>1</sup>+5×3<sup>2</sup>+5×3<sup>3</sup> = 202

We see a pattern and we would suppose that the next term is

6th term = 2+5×3<sup>0</sup>+5×3<sup>1</sup>+5×3<sup>2</sup>+5×3<sup>3</sup>+5×3<sup>4</sup> = 607

So we assume that the general term is:

kth term = 2+5×3<sup>0</sup>+5×3<sup>1</sup>+5×3<sup>2</sup>+···+5×3<sup>k-2</sup>

or factoring 5 out of all those with factor 5:

kth term = 2+5(3<sup>0</sup>+3<sup>1</sup>+3<sup>2</sup>+···+3<sup>k-2</sup>)

The terms in parentheses is the sum of a geometric sequence with 
{{{a[1]}}}=1, {{{a[n]}}}=3<sup>k-2</sup>,  r=3,   

S<sub>n</sub> = {{{(a[1]-r*a[n])/(1-r)}}} 

{{{(1-3*3^(k-2))/(1-3)}}} = {{{(1-3^(k-1))/(-2)}}} = {{{(-(-1+3^(k-1)))/(-2)}}} = {{{(-1+3^(k-1))/2}}} = {{{(3^(k-1)-1)/2}}}

Now let's go back the the general term 

kth term = 2+5(3<sup>0</sup>+3<sup>1</sup>+3<sup>2</sup>+···+3<sup>k-2</sup>)

and substitute for the terms in the parentheses:

kth term = 2+5{{{((3^(k-1)-1)/2)}}} = {{{2/1+expr(5/1)((3^(k-1)-1)/2)}}} = {{{4/2+(5(3^(k-1)-1))/2)}}} = 
{{{4/2+(5*3^(k-1)-5)/2)}}} = {{{(4+5*3^(k-1)-5)/2)}}} = {{{(5*3^(k-1)-1)/2)}}}

So if you want to call it the nth term instead of the kth term, just
use n instead of k:

{{{a[n]}}}{{{""=""}}}{{{(5*3^(n-1)-1)/2)}}}

You can crank out as many terms as you like.  Here are the first 20:

2, 7, 22, 67, 202, 607, 1822, 5467, 16402, 49207, 147622, 442867, 1328602,
3985807, 11957422, 35872267, 107616802, 322850407, 968551222, 2905653667.

Edwin</pre>