Math
- Thread starter INAARA
- Start date
Have you seen the "Read before posting" section? What course are you in? What are you working on right now?
Do you know what cubic equations are? If the first column is x and the second column is y, you might look for something in the form
y = Ax^3 + Bx^2 +Cx + D
You have 8 data points, but you only need 4 to set up the four equations necessary to solve for A, B, C, and D.
Sorry.
Soorry... I forgot
and im in grade 8
and i'm learning algebraic expression
in math
n Yes i have heard of cubic expressions
but i need to find a algebraic expression to the term number (first column) to get to the term value (second column)
and i dont think what you said was right the ABCD thing
because i was looking for something more like n+2-9 or something with powers
Soorry... I forgot
and im in grade 8
and i'm learning algebraic expression
in math
n Yes i have heard of cubic expressions
but i need to find a algebraic expression to the term number (first column) to get to the term value (second column)
and i dont think what you said was right the ABCD thing
because i was looking for something more like n+2-9 or something with powers
Hello, INAARA!
This is quite an advanced problem for 8th grade.
It requires some techniques not taught in Algebra I.
Take the differences of cosecutive terms,
. . then take the differences of the differences, and so on.
. . \(\displaystyle \begin{array}{cccccccccccccc}\text{Sequence} & 1 && 4 && 10 && 20 && 35 && 56 \\ \text{1st diff.} && 3 && 6 && 10 && 15 && 21 \\ \text{2nd diff.} &&& 3 && 4 && 5 && 6 \\ \text{3rd diff.} &&&& 1 && 1 && 1 \end{array}\)
We see that the third differences are constant.
Hence, the generating function is of the third degree . . . a cubic.
The general cubic function is:.\(\displaystyle f(x) \:=\:an^3 + bn^2 + cn + d\)
. . and we must determine the values of \(\displaystyle a,b,c,d.\)
Use the first four values to set up a system of equations:
. . \(\displaystyle \begin{array}{cccccccccc}f(1) = 1\!: & a + b + c + d &=& 1 & [1] \\ f(2) = 4\!: & 8a + 4b + 2c + d &=& 4 & [2] \\ f(3) = 10\!: & 27a + 9b + 3c + d &=& 10 & [3] \\ f(4) = 20\!: & 64a + 16b + 4c + d &=& 20 & [4] \end{array}\)
Solve the system and get: .\(\displaystyle a = \frac{1}{6},\;b = \frac{1}{2},\;c = \frac{1}{3},\;d = 0\)
Hence:.\(\displaystyle f(n) \:=\:\frac{1}{6}n^3 + \frac{1}{2}n^2 + \frac{1}{3}n \:=\;\frac{n}{6}(n^2+3n+2)\)
Therefore: .\(\displaystyle f(n) \;=\;\dfrac{n(n+1)(n+2)}{6}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
An interesting note . . .
In the 1st differences, I recognized the "triangular numbers":
. . \(\displaystyle 1,3,6,10,15, \hdots\)
You can see why they are called "triangular".
. . \(\displaystyle \begin{array}{ccccccc}&&&&&& \bullet \\ &&&& \bullet && \bullet\;\bullet \\ && \bullet && \bullet\;\bullet && \bullet\;\bullet\;\bullet \\ \bullet && \bullet\;\bullet && \bullet\;\bullet\;\bullet && \bullet\;\bullet\;\bullet\;\bullet \\ 1 && 3 && 6 && 10\end{array}\)
So the sequence is the sum of consecutive triangular numbers.
And I happened to know the formula for that.
. . so I solved the problem by "eyeballing" it.
Here's the "interesting" part . . .
The triangular numbers are part of "The Twelve Days of Chrismas".
On the 1st day of Christmas, my true love gave to me: 1 gift.
On the 2nd day of Christmas, she gave me: 1 + 2 gifts.
On the 3rd day of Christmas, she gave me: 1 + 2 + 3 gifts.
. . and so on.
Question: How many gifts did my true love give to me during the 12 days?
Answer:. \(\displaystyle f(12) \;=\;\dfrac{(12)(13)(14)}{6} \;=\;364\text{ gifts.}\)
. . ta-DAA!
This is quite an advanced problem for 8th grade.
It requires some techniques not taught in Algebra I.
Take the differences of cosecutive terms,
. . then take the differences of the differences, and so on.
. . \(\displaystyle \begin{array}{cccccccccccccc}\text{Sequence} & 1 && 4 && 10 && 20 && 35 && 56 \\ \text{1st diff.} && 3 && 6 && 10 && 15 && 21 \\ \text{2nd diff.} &&& 3 && 4 && 5 && 6 \\ \text{3rd diff.} &&&& 1 && 1 && 1 \end{array}\)
We see that the third differences are constant.
Hence, the generating function is of the third degree . . . a cubic.
The general cubic function is:.\(\displaystyle f(x) \:=\:an^3 + bn^2 + cn + d\)
. . and we must determine the values of \(\displaystyle a,b,c,d.\)
Use the first four values to set up a system of equations:
. . \(\displaystyle \begin{array}{cccccccccc}f(1) = 1\!: & a + b + c + d &=& 1 & [1] \\ f(2) = 4\!: & 8a + 4b + 2c + d &=& 4 & [2] \\ f(3) = 10\!: & 27a + 9b + 3c + d &=& 10 & [3] \\ f(4) = 20\!: & 64a + 16b + 4c + d &=& 20 & [4] \end{array}\)
Solve the system and get: .\(\displaystyle a = \frac{1}{6},\;b = \frac{1}{2},\;c = \frac{1}{3},\;d = 0\)
Hence:.\(\displaystyle f(n) \:=\:\frac{1}{6}n^3 + \frac{1}{2}n^2 + \frac{1}{3}n \:=\;\frac{n}{6}(n^2+3n+2)\)
Therefore: .\(\displaystyle f(n) \;=\;\dfrac{n(n+1)(n+2)}{6}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
An interesting note . . .
In the 1st differences, I recognized the "triangular numbers":
. . \(\displaystyle 1,3,6,10,15, \hdots\)
You can see why they are called "triangular".
. . \(\displaystyle \begin{array}{ccccccc}&&&&&& \bullet \\ &&&& \bullet && \bullet\;\bullet \\ && \bullet && \bullet\;\bullet && \bullet\;\bullet\;\bullet \\ \bullet && \bullet\;\bullet && \bullet\;\bullet\;\bullet && \bullet\;\bullet\;\bullet\;\bullet \\ 1 && 3 && 6 && 10\end{array}\)
So the sequence is the sum of consecutive triangular numbers.
And I happened to know the formula for that.
. . so I solved the problem by "eyeballing" it.
Here's the "interesting" part . . .
The triangular numbers are part of "The Twelve Days of Chrismas".
On the 1st day of Christmas, my true love gave to me: 1 gift.
On the 2nd day of Christmas, she gave me: 1 + 2 gifts.
On the 3rd day of Christmas, she gave me: 1 + 2 + 3 gifts.
. . and so on.
Question: How many gifts did my true love give to me during the 12 days?
Answer:. \(\displaystyle f(12) \;=\;\dfrac{(12)(13)(14)}{6} \;=\;364\text{ gifts.}\)
. . ta-DAA!