# Pattern question is stumping me!

• Oct 20th 2010, 09:35 AM
squall458
Pattern question is stumping me!
-3, -2, 0, 3

Can someone tell me what the algebra expression is for this pattern. One variable only. Thanks so much!!!
• Oct 20th 2010, 10:09 AM
masters
Quote:

Originally Posted by squall458
-3, -2, 0, 3

Can someone tell me what the algebra expression is for this pattern. One variable only. Thanks so much!!!

Hi squall458,

I think Soroban will come up with an eloquent explanation. I just ran a quadratic regression on my calculator for the sequence:

Code:

X:  1    2    3    4    5    6    7 Y: -3  -2    0    3    7    12  18
And came up with $\displaystyle y=.5x^2-.5x-3$
• Oct 20th 2010, 10:15 AM
Soroban
Hello, squall458!

I'll show you a procedure for this problem.
It's rather long, but it's simple and direct.

Quote:

$\displaystyle -3, -2, 0, 3$

Can someone tell me what the algebra expression is for this pattern?

Take the difference of consecutive terms,
. . then take difference of the differences and so on.

. . $\displaystyle \begin{array}{cccccccccc} \text{Sequence} & \text{-}3 && \text{-}2 && 0 && 3 & \hdots \\ \text{1st difference} && 1 && 2 && 3 \\ \text{2nd difference} &&& 1 && 1 \end{array}$

We see that the second differences are constant.
Hence, 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 and set up a system of equations:

. . $\displaystyle \begin{array}{cccccc} f(1) = \text{-}3: & a + b + c &=& \text{-}3 & [1] \\ f(2) = \text{-}2: & 4a + 2b + c &=& \text{-}2 & [2] \\ f(3) = 0: & 9a + 3b + c &=& 0 & [3] \end{array}$

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

. . $\displaystyle \begin{array}{cccccccc} \text{Subtract [5] - [4]:} & 2a \:=\: 1 & \Rightarrow & a \:=\:\frac{1}{2} \end{array}$

. . $\displaystyle \begin{array}{ccccccc}\text{Substitute into [4]:} & 3(\frac{1}{2}) + b \:=\:1 & \Rightarrow & b \:=\:\text{-}\frac{1}{2} \end{array}$

. . $\displaystyle \begin{array}{ccccccc}\text{Substitute into [1]:} & \frac{1}{2} - \frac{1}{2} + c \:=\:\text{-}3 & \Rightarrow & c \:=\:\text{-}3 \end{array}$

Therefore, the generating function is:

. . . $\displaystyle f(n) \;=\;\frac{1}{2}n^2 - \frac{1}{2}n - 3 \;=\;\dfrac{(n-3)(n+2)}{2}$
• Oct 20th 2010, 11:11 AM
masters
Told you he would!!