# Thread: Problem solving - sucessive tetrahedral numbers

1. ## Problem solving - sucessive tetrahedral numbers

would you guys be able to give me a hand with this problem solving question, i have no idea where to start...

Thanks

Zac

2. Hello, Zac!

12. The first four tetrahedral numbers are: . $1,\;4,\;10,\;20$

Find the pattern, then predict the next 3 terms of the sequence.
Take the difference of consecutive pairs of terms.

$\begin{array}{cccccccc}\text{Sequence:} & 1 && 4 && 10 && 20 \\
\text{Difference:} & & 3 & & 6 & & 10 \end{array}$

They are adding on consecutive "triangular" numbers.

. . $\begin{array}{ccc}3 & = & 1+2 \\ 6 &=&1+2+3\\ 10 &=&1+2+3+4 \end{array}$

We will add on the next three triangular numbers:

. . $\begin{array}{ccc}1+2+3+4+5 &=& 15 \\ 1+2+3+4+5+6 &=& 21 \\ 1+2+3+4+5+6+7 &=& 28\end{array}$

$\begin{array}{cccccccccccccc}\text{Sequence:} & 1 && 4 && 10 && 20 && {\color{blue}35} && {\color{blue}56} && {\color{blue}84}\\
\text{Difference:} & & 3 & & 6 & & 10 && +15 && +21 && +28\end{array}$

What is the $n^{th}$ term of the sequence?
I'll skip all the algebra . . .

$\text{The }n^{th}\text{ term is: }\;a_n \;=\;\frac{n(n+1)(n+2)}{6}$

3. Originally Posted by Soroban

I'll skip all the algebra . . .

$\text{The }n^{th}\text{ term is: }\;a_n \;=\;\frac{n(n+1)(n+2)}{6}$
as this is a problwm solving questions and according to my sheet i must show the algebra would you mind putting it up?

thanks so much!

Zac

4. Hello, Zac!

This will take a while . . . better sit down.

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

$\begin{array}{cccccccccccccc}\text{Sequence} & 1 && 4 && 10 && 20 && 35 && 56 && 84 \\
\text{1st diff.} & & 3 && 6 && 10 && 15 && 21 && 28 \\
\text{2nd diff.} & & & 3 && 4 && 5 && 6 && 7 \\
\text{3rd diff.} & & & & 1 && 1 && 1 && 1 \end{array}$

The third differences are constant.
. . This tells us that the generating fuction is a cubic.

The general cubic function is: . $f(n) \;=\;an^3 + bn^2 + cn + d$

We will use the first four terms of the sequence . . .

$\begin{array}{ccccc}f(1)\:=\:1\!: & a + b + c + d & = & 1 & [1] \\
f(2) \:=\:4\!: & 8a + 4b + 2c + d &=& 4 & [2] \\
f(3) \:=\:10\!: & 27a + 9b + 3c + d &=& 10 & [3] \\
f(4) \:=\:20\!: & 64a + 16b + 4c + d &=& 20 & [4] \end{array}$

$\begin{array}{ccccc}\text{Subtract [2] - [1]:} & 7a + 3b + c &=& 3 & [5] \\
\text{Subtract [3] - [2]:} & 19a + 5b + c &=& 6 & [6] \\
\text{Subtract [4] - [3]:} & 37a + 7b + c &=& 10 & [7] \end{array}$

$\begin{array}{ccccc}\text{Subtract [6] - [5]:} & 12a + 2b &=& 3 & [8] \\
\text{Subtract [7] - [6]:} & 18a + 2b &=& 4 & [9] \end{array}$

$\begin{array}{ccccccc}\text{Subtract [9] - [8]:}& 6a \;=\; 1 & \Rightarrow & \boxed{a \;=\;\frac{1}{6}} \end{array}$

Substitute into [8]: . $12\left(\frac{1}{6}\right) + 2b \:=\:3\quad\Rightarrow\quad\boxed{ b \:=\:\frac{1}{2}}$

Substitute into [5]: . $7\left(\frac{1}{6}\right) + 3\left(\frac{1}{2}\right) + c \:=\:3 \quad\Rightarrow\quad \boxed{c\:=\:\frac{1}{3}}$

Substitute into [1]: . $\frac{1}{6} + \frac{1}{2} + \frac{1}{3} + d \:=\:1 \quad\Rightarrow\quad\boxed{ d\:=\:0}$

Hence, the function is: . $f(n) \;=\;\frac{1}{6}n^3 + \frac{1}{2}n^2 + \frac{1}{3}n \;=\;\frac{n}{6}(n^2+3n+ 2)$

. . Therefore: . $\boxed{f(n) \;=\;\frac{n(n+1)(n+2)}{6}}$