<s>[INST] finding nth term Let 
  3,8,17,32,57 . . . . .
be a pattern.How do we find the nth number?My brains are completely jammed,I am tired.I do not even recognize the pattern.I calculated a few ways,but all I want is a little hint,not the whole solution.
 [/INST] Note: I just added a derivation of an explicit formula
for the terms as the OP requested.
From 3,8,17,32,57,
if the term is n,
the next term is 2n+k
where k = 2, 1, -2, -7.
The differences of k are
-1, -3, -5.
If we assume that the next difference is -7,
the next k is -7-7=-14
and the next term is 2*57-14 = 100.
To get a formula from this,
since the sum of the odd numbers are the squares,
the values of k are $2-m^2$
starting with $m = 0$.
Letting $s(0) = 3$,
$s(n+1) = 2s(n)+2-n^2$.
This is made explicit below.
Check:
$s(1) = 2*3+2-0 = 8$,
$s(2) = 2*8+2-1 = 17$,
$s(3) = 2*17+2-4 = 32$,
$s(4) = 2*32+2-9 = 57$.
To get an explicit form for
$s(0) = 3$,
$s(n+1) = 2s(n)+2-n^2$
let $s(n) = 2^n t(n)$.
Then $2^{n+1}t(n+1) = 2^{n+1}t(n)+2-n^2$,
or
$t(n+1) = t(n) + 1/2^{n}-n^2/2^n$,
so
$t(n+1) - t(n) = 1/2^{n}-n^2/2^n$.
Summing this,
$t(m)-t(0) = \sum_{n=0}^{m-1} (t(n+1) - t(n))
= \sum_{n=0}^{m-1} (1/2^{n}-n^2/2^n)
= 1-1/2^m - \sum_{n=0}^{m-1}n^2/2^n
$.
To evaluate the last sum,
let $f(x) = \sum_{n=0}^{m-1} x^n
= (1-x^m)/(1-x)
$.
Then $x f'(x) = \sum_{n=0}^{m-1}  nx^n$
and $x (x f'(x))' = \sum_{n=0}^{m-1}  n^2 x^n$.
We want $x (x f'(x))'$ at $x = 1/2$.
Note that $x (x f'(x))'
= x (f'(x) + x f''(x))
= x f'(x) + x^2 f''(x)
$.
I'm feeling lazy, so I won't bother getting
$f'$ and $f''$.
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