Question 1107256
<pre>
n   {{{a[n]}}}    diff      diff(diff)
1     3                 
2     5       2                    
3    11       6           4
4    21      10           4

</pre>

So if we assume  {{{ a[n] = k[1]n^2 + k[2]n + k[3] }}}
we can write
(1)  {{{ 3 = k[1] + k[2] + k[3] }}}
(2)  {{{ 5 = k[1](2^2) + k[2](2) + k[3] }}}  
(3)  {{{ 11 = k[1](3^2) + k[2](3) + k[3] }}}

This system can be solved to get {{{ k[1] = 2 }}}, {{{ k[2] = -4 }}}, and {{{ k[3] = 5 }}}

giving  {{{ a[n] = 2n^2 -4n + 5 }}};  n = 1,2,3,…

and {{{ highlight( a[48] = 4421 ) }}}

——

There is probably a method to solve this using z-transforms but I'm very rusty on them.