InOut Table

I really need help to figure out this In Out table.
I've been trying for a few minutes now, it puzzles me, any help?

IN ....... OUT
-------------------
2 ...........5
3 ...........14
4.............26
5.............42

Get it?

Hello, CompletelyConfused!

This is not a simple proboem . . .

I really need help to figure out this In-Out table.

. . \(\displaystyle \begin{array}{c|c} \text{IN} & \text{OUT} \\ \hline 2 & 5 \\ 3 & 14 \\ 4 & 26 \\ 5 & 42 \\ 6 & 83 \end{array}\)

Click to expand...

As Denis suggested, look at the differences of consecutive terms,
. . then the differences of the differences, and so on.

. . \(\displaystyle \begin{array}{c|ccccccccc} \hline n & 2 && 3 && 4 && 5 && 6\\ \hline \hline f(n) & 5 && 14 && 26 && 42 && 63 \\ \hline \text{1st diff} && 0 && 12 && 16 && 21 \\ \text{2nd diff} &&& 3 && 4 && 5 \\ \text{3rd diff} &&&&1 && 1 \end{array}\)

The third differences are constant.
This tells us that the generating function is of the third degree: a cubic.


The general cubic function is: .\(\displaystyle f(n) \;=\;an^3 + bn^2 + cn + d\)
. . and we must determine \(\displaystyle a,b,c,d.\)


Use the first 4 values of the sequence:

. . \(\displaystyle \begin{array}{cccccc} f(2) = 5\; & 8a + 4b + 2c + d &=& 5 \\ f(3) = 14 & 27a + 9b + 3c + d &=& 14 \\ f(4) = 26 & 64a + 16b + 4c + d &=& 26 \\ f(5) = 42 & 125a + 25b + 5c + d &=& 42 \end{array}\)

Then solve the system of equations.


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

\(\displaystyle \text{I got: }\;a = \tfrac{1}{6},\;b = 0,\;c = \tfrac{35}{6},\;d = -8\)

\(\displaystyle \text{Therefore: }\;f(n) \;=\;\frac{1}{6}n^3 + \frac{35}{6}n - 8 \;=\;\frac{1}{6}(n^3 + 35n - 48)\)