Question 512608
<pre>
          1st          2nd  
n  a<sub>n</sub>  differences differences
1   5     0           1               
2   5     1           1
3   6     2           1  
4   8     3           1  
5  11     4
6  15     

Since the 2nd differences are all constant we try a 2nd
degree polynomial in n

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

{{{system(5=a[1] = A(1)^2+B(1)+C, 
          5=a[2] = A(2)^2+B(2)+C, 
          6=a[3] = A(3)^2+B(3)+C)}}}

Simplifying:

{{{system(5= A(1)+B+C, 
          5= A(4)+2B+C, 
          6= A(9)+3B+C)}}}{{{system(5= A+B+C, 
          5= 4A+2B+C, 
          6= 9A+3B+C)}}}{{{system(A+B+C=5, 
          4A+2B+C=5, 
          9A+3B+C=6)}}}

Solve that system and get A = {{{1/2}}}, B = {{{-3/2}}}, C = 6

So the equation

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

becomes:

{{{a[n] = expr(1/2)n^2-expr(3/2)n+6}}}

{{{a[n] = expr(1/2)n^2-expr(3/2)n+12/2}}}

{{{a[n] = (n^2-3n+12)/2}}}

Substituting n = 20

{{{a[20] = (20^2-3*20+12)/2}}}

{{{a[20] = (400-60+12)/2}}}

{{{a[20] = 352/2}}}

{{{a[20] = 176}}}

Substituting n = 21

{{{a[21] = (21^2-3*21+12)/2}}}

{{{a[21] = (441-63+12)/2}}}

{{{a[21] = 390/2}}}

{{{a[21] = 195}}}


Answers: {{{a[20] = 176}}}, {{{a[21] = 195}}}

Edwin</pre>