The document describes a tribonacci sequence and how to find the nth term of the sequence. It provides the terms up to the 13th term and asks to find the next three terms, which are 193, 355, 653. It explains that each term is the sum of the preceding three terms. The document then discusses how to represent this sequence mathematically using the recurrence formula Tn=Tn-3+Tn-2+Tn-1 to find future terms of the sequence. It also explains how to use this formula to find the number of tiles needed to construct the next figure in a sequence of square tile patterns.

1. Investigate:1 1 1 3 5 9 17 31 57 105
a. Find the next three terms:
b. Write a mathematical sentence
which can be used as the formula in
finding the nth term of the sequence.
1 1 1 3 5 9 17 31 57 105 193 355 653
a. The next three terms are as
follows, 193 for the 11th term, 355 for
the 12th term, and 653 for the 13th term.
b. If we are going to investigate this
3
9
31
105
193
355
653
11th 12th 13th
5
17
57

2. sequence, we can observe that for each
number in the series (starting from the
value 3 [4th term] to last) is the sum of
preceding 3 numbers.
We can deduce that this is a Tribonacci
Sequence. The Tribonacci sequence
starts with three predetermined terms
and then each successive term is the
sum of the three preceding terms.
The table below shows the observation:
What to find (value) How to find
4th
term (3) 1st
+ 2nd
+ 3rd
terms (1+1+1)
5th
term (5) 2nd
+ 3rd
+ 4th
terms (1+1+3)
6th
term (9) 3rd
+ 4th
+ 5th
terms (1+3+5)
7th
term (17) 4th
+ 5th
+ 6th
terms (3+5+9)

3. In order to solve for the next three
terms (11th, 12th and 13th), we will make
use of this recurrence equation on
finding for the nth term where n ≥ 4:
Example, this is the formula if we want
to find the 4th term of a Tribonacci
sequence.
T4=T1+T2+T3 which comes from the
equation
Tn= Tn−3+ Tn−2+Tn−1 (the formula for
8th
term (31) 5th
+ 6th
+ 7th
terms (5+9+17)
9th
term(57) 6th
+ 7th
+8th
terms (9+17+31)
10th
term(105) 7th
+8th
+9th
terms (17+31+57)
11th
term (193) 8th
+9th
+ 10th
terms (31+57+105)
12th
term (355) 9th
+ 10th
+ 11th
terms (57+105+193)
13th
term (653) 10th
+11th
+12th
terms (105+193+355)

4. finding the nth term of a tribonacci
sequence)

5. 3. For this kind of tile sequence, we can assert
that we can have different approach for solving the
nth figure. In our way, we used a function and
figures to solve this kind of problem.
First, we need to fill out each figures as a dimension
of square. To do this, let us name first the shown
sequence of tiles as figures
Visualizing the above sentence, here are the figures
formed:
Figure 1 Figure 2 Figure 3
2x2
dimension
3x3
dimension
4x4
dimension

6. Figure 4
Knowing that every nxn dimension are added by
1x1 dimension to arrive for the next figure, we can
conclude that the next figure would be a 6x6
dimension square. We can also explore from the
figures that in order to arrive at the dimension, we
can show it on mathematical expression (n+2)(n+2).
Say, for example, for figure 1, we noticed that we
5x5
dimension

7. don’t need to substitute any natural numbers to get
2x2 dimension for the first figure. Therefore, with a
view to get 2x2 dimension, we will just substitute
zero on the expression. (0+2)(0+2) will be simplified
to (2)(2) or 2x2 dimension. Now for the figure 2, we
noticed that the tiles are in 3x3 form, therefore, in
order to arrive for 3x3 dimension, we are going to
add 1 shift horizontally and 1 shift vertically which
then can be substituted to the expression which can
be seen as (1+2)(1+2) => (3)(3) or 3x3 dimension.
The pattern which can be noticed could be traced
for the next figures. Now, we can see that the tiles
without shades can be expressed into mathematical
expression 2(n+1), (n+1) tiles for the first column
and (n+1) tiles for the last column. To better
understand, we can visually show the expression
using the figures above. Say for example, for figure

8. 3:
We arrived at the expression 2(n+1) to get the total
number of tiles which are not shaded by blue.
If we are going to construct a table for better
understanding, follow below.
n+1
2+1
3 non-shaded tiles
for first column n+1
2+1
3 non-shaded tiles
for last column
2(2+1) = 4+2 = 6 non-
shaded tiles

9. Now, going back to the problem letter a. How many
square tiles will it take to build the next figure?
Since we know that the predicted dimension of the
figure (which we’ll be calling figure 5) is 6x6. We can
tell that there are 4 shifts added vertically and
Figure Dimension Shift added
vertically
and
horizontally
(n)
Mathematical
expression
(n+2)(n+2)
Mathematical
expression
2(n+1)
1 2x2 0 shift /
No shift
(0+2)(0+2) 2(0+1)
2 3x3 1 shift (1+2)(1+2) 2(1+1)
3 4x4 2 shifts (2+2)(2+2) 2(2+1)
4 5x5 3 shifts (3+2)(3+2) 2(3+1)
5 6x6 4 shifts (4+2)(4+2) 2(4+1)Next figure

10. horizontally inferring to the figure 1 which can be
shown below.
Now in, order to find the the non-shaded
tiles, we will just substitute n=4 to 2(n+1) =>
2(4+1) = 10 non-shaded tiles.
Added shifts four
units horizontally
Added shifts four
units vertically
6x6 dimension square

11. This basically means that there are 5 non-
shaded tiles for top left column and 5 non-
shaded tiles for bottom right column of a 6x6
dimension which can be shown below.
We are asked to find the number of square
tiles (shaded tiles).
We can find the shaded tiles by inspection which is
26 shaded tiles. On the other side, we can also
figure out the number of the shaded tiles by making
a function shown below.
n+1
4+1
5 non-shaded tiles
for last column
n+1
4+1
5 non-shaded tiles
for first column

12. f(n) = (n+2)(n+2)-2(n+1)
where (n+2)(n+2) will be the
arrangement/dimension of the tiles and 2(n+1) are
tiles which are non-shaded.
Knowing that n=4, we can substitute it to the
function;
f(4) = (4+2)(4+2) – 2(4-1)
= (6)(6) – 2(5)
= 36 – 10
= 26
Therefore, we conclude that we can also get the
number of shaded tiles by using functions.
b. To get the number of tiles for an nth figure,
we will subtract 2(n+1) from (n+2)(n+2) which can
be shown as a function:
Solving by Function =
Solving by Inspection
26 = 26
TRUE

13. f(n) = (n+2)(n+2)-2(n+1)