Question 929822
<pre>
Put them in a column.  Out beside each number write what
you get whe you subtract each number from the one just below it.
Then make a third column doing that with the second column

 1   1   1 
 2   2   1
 4   3   1
 7   4   1
11   5   1
16   5
22

Those columns are called the first and second differences.

Since it takes 2 columns of differences to get a column that has 
all the same number, we assume a polynomial of degree 2 for the 
nth term {{{t[n]}}}.

{{{t[n] = An^2+Bn+C}}}

Substitute n=1

{{{t[1] = A(1)^2+B(1)+C}}}

and since the first term is 1, we can write

{{{1 = A+B+C}}} or

{{{A+B+C=1}}}

Substitute n=2

{{{t[2] = A(2)^2+B(2)+C}}}

and since the second term is 2, we can write

{{{2 = 4A+2B+C}}} or

{{{4A+2B+C=2}}}

Substitute n=3

{{{t[3] = A(3)^2+B(3)+C}}}

and since the third term is 4, we can write

{{{4 = 9A+3B+C}}} or

{{{9A+3B+C=4}}}

Now we have a system of equations

{{{system(A+B+C=1,4A+2B+C=2,9A+3B+C=4)}}}

Solve that system and get {{{A=1/2}}},{{{B=-1/2}}},{{{C=1}}}

So the nth term,

{{{t[n] = An^2+Bn+C}}}  

becomes

{{{t[n] = expr(1/2)n^2-expr(1/2)n+1}}}

or if you prefer you can get an LCD

{{{t[n] = (n^2-n+2)/2}}}

Then for the 16th term, we substitute n=16

{{{t[n] = (16^2-16+2)/2}}} = {{{(256-16+2)/2}}} = {{{121}}}

Checking:

1,2,4,7,11,16,22,29,37,46,56,67,79,92,106,121

Edwin</pre>