In and Out table

[laurienice]

Hello, i've been working on this In and Out table for about 3 hours now, and i can't fiugre it out. Can anyone help me? Please?!


IN OUT
3 0
4 2
5 5
6 9
7 14

I need to find a rule for this table, and i can't anyone want to help

-Cassidy.

 

[Deleted member 4993]

Hello, i've been working on this In and Out table for about 3 hours now, and i can't fiugre it out. Can anyone help me? Please?!


IN OUT
3 0
4 2
5 5
6 9
7 14

I need to find a rule for this table, and i can't anyone want to help

-Cassidy.

Note that the output is increasing by 2,3,4,5......

 

[Deleted member 4993]

What level of mathematics are you doing - which grade?

Please show some of your work - so that I know where to begin.

 

[soroban]

Hello, laurienice!

Find the rule for this table.

.\(\displaystyle \begin{array}{|c|c|}\hline \text{In } & \text{Out} \\ \hline 3 & 0 \\ 4 & 2 \\ 5 & 5 \\ 6 & 9 \\ 7 & 14 \\ \vdots & \vdots \end{array}\)

Take the difference of consecutive terms . . . Then take the difference of the differences, etc.

. . \(\displaystyle \begin{array}{cccccccccc} \text{Sequence} & 0 && 2 && 5 && 9 && 14 \\ \text{1st diff} && 2 && 3 && 4 && 5 \\ \text{2nd diff} & & & 1 && 1 && 1 \end{array}\)

The second differences are constant.
This tell us that the generating function is of the second degree . . . a quadratic.


The general quadratic function is: .\(\displaystyle f(n) \:=\:an^2 + bn + c\)

Use the first three terms of the sequence to determine a, b, c.

. . \(\displaystyle \begin{array}{ccccc}f(3) \:=\:0 & 9a + 3b + c &=& 0 & [1] \\ f(4) \:=\:2 & 16a + 4b + c &=& 2 & [2] \\ f(5) \:=\:5 & 25a + 5b + c &=& 5 & [3] \end{array}\)

\(\displaystyle \begin{array}{ccccc}\text{Subtract [2] - [1]:} & 7a + b &=& 2 & [4] \\ \text{Subtract [3] - [2]:} & 9a + b &=& 3 & [5] \end{array}\)

\(\displaystyle \text{Subtract [5] - [4]: }\:2a \:=\:1 \quad\Rightarrow\quad \boxed{a \:=\:\tfrac{1}{2}}\)

\(\displaystyle \text{Substitute into [4]: }\;\tfrac{7}{2} + b \:=\:2 \quad\Rightarrow\quad \boxed{b \:=\:-\tfrac{3}{2}}\)

\(\displaystyle \text{Substitute into [1]: }\:\tfrac{9}{2} - \tfrac{9}{2} + c \:=\:0 \quad\Rightarrow\quad\boxed{ c \:=\:0}\)


\(\displaystyle \text{Therefore: }\;f(n) \;=\;\tfrac{1}{2}n^2 - \tfrac{3}{2}n \;=\;\frac{n(n-3)}{2}\)

 

[laurienice]

It's 9th Grade