finding nth term

Question

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.

Answer 1

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 $f''$.

Answer 2

Hint: One solution can be obtained by taking differences-of-differences.

Answer 3

If you want the smallest degree polynomial that gives these numbers (which is only one of the infinitely many possible explanations for them), we can use the calculus of finite differences: $$\fbox{3}\quad 8\quad 17\quad 32\quad 57 \\ \fbox{5}\quad 9\quad 15\quad 25\\ \fbox{4}\quad 6\quad 10\\ \fbox{2} \quad 4\\ \fbox{2}$$ Define $$\begin{align*} f(n)&=3\binom{n}{0} +5\binom{n}{1}+4\binom{n}{2}+2\binom{n}{3}+2\binom{n}{4}\\\\ &=3 + \frac{19 n}{6} + \frac{23 n^2}{12} - \frac{n^3}{6} + \frac{n^4}{12} \end{align*}$$ Then $$f(0)=3,\quad f(1)=8,\quad f(2)=17,\quad f(3)=32,\quad f(4)=57$$