Question 822545
<pre>
16,49,104,181

I always first check the sequence of differences to see
if I can recognize a pattern:

2nd term - 1st term =   49-16 = 33 = 11×3
3rd term - 2nd term =  104-49 = 55 = 11×5 
4th term - 3rd term = 181-104 = 77 = 11×7

So we can now see the pattern.

 1.    Start with 16, then
 2.    16 + 11×3  =   49

and the multiple of 11, which is 3, is 1 less than twice the term
number, which is 2. 

 3.    49 + 11×5  =  104 =

16 + 11×3 + 11×5

and the multiple of 11, which is 5, is 1 less than twice the term
number, which is 3.


 4.   104 + 11×7  =  181 =

16 + 11×3 + 11×5 + 11×7

and the multiple of 11, which is 7, is 1 less than twice the term
number, which is 4

Therefore to find the 20th term, we start with 16 and add
19 terms.  So the 20th term is

16+{{{sum(11(2k-1),k=2,20)}}} =

Factor out 11 from the sun:

16+{{{11*sum((2k-1),k=2,20)}}} =

Break up the sum into two sums:

16+{{{11*(sum((2k),k=2,20)-sum((1),k=2,20))  }}} =

Factor out 2 from the first sum:

16+{{{11*(2*sum((k),k=2,20)-sum((1),k=2,20))  }}} =

The sum {{{sum((k),k=2,20)}}} is the sum of an arithmetic sequence
with a<sub>1</sub> = 2 and a<sub>19</sub> = 20, and d=1

S<sub>n</sub> = {{{n/2}}}(a<sub>1</sub> + a<sub>n</sub>)
S<sub>19</sub> = {{{19/2}}}(2 + 20) = {{{19/2}}}(22) = 209

And {{{sum((1),k=2,20))}}} is the sum of 19 ones, which is 19.

Substituting in:

16+{{{11*(2*sum((k),k=2,20)-sum((1),k=2,20))  }}} =

We have

16+11·(2·209-19) =

4405.

Edwin</pre>