Question 983310
<pre>
A very common recursion formula is of this form:

{{{t[n+1]= a*n*t[n]+b*n+c}}}, where a,b, and c are constants:

Substitute n=1

{{{t[1+1]= a*1*t[1]+b*1+c}}}
{{{t[2]= a*1*1+b+c}}}
{{{2= a+b+c}}}

Substitute n=2

{{{t[2+1]= a*2*t[2]+b*2+c}}}
{{{t[3]= 2a*14+2b+c}}}
{{{24= 28a+2b+c}}}

Substitute n=3

{{{t[3+1]= a*3*t[3]+b*3+c}}}
{{{t[4]= 3a*24+3b+c}}}
{{{66= 72a+3b+c}}}

We will find a,b, and c from this system:

{{{system(16a+b+c=14,28a+2b+c=24,72a+3b+c-66)}}}

Subtracting the 1st equation from the 2nd equation

{{{12a+b=10}}}

Subtracting the 2nd equation from the 3rd equation

{{{44a+b=42}}}

So now we have this system:

{{{system(12a+b=10,44a+b=42)}}}

Subtracting the first equation from the second:

{{{32a=32}}}

{{{a=1}}}

Substitute in 12a+b=10

12(1)+b=10
   12+b=10
      b=-2

Substitute in 16a+b+c=14

16(1)+(-2)+c=14
      16-2+c=14
        14+c=14
           c=0

So recursion formula:

{{{t[n+1]= 1*n*t[n]+(-2)*n+0}}}

{{{t[n+1]= n*t[n]-2n}}}

Now we test the recursion formula to see if {{{t[5]=256}}}

{{{t[4+1]= 4*t[4]-2*4}}}

{{{t[5]= 4*66-8}}}

{{{t[5]= 256}}}

So this is the correct recursion formula.

So we use it to find {{{t[6]}}}

{{{t[5+1]= 5*t[5]-2*5}}}

{{{t[6]= 5*256-10}}}

{{{t[6]= 1270}}}   <--answer

Edwin</pre>