Exercise 69 Page 450

Practice makes perfect

a First, let's analyze the sequence

T_{n} .

Triangular Numbers

The numbers

T_{n}

are represented by the points of following diagrams.

Note that the

n^{th}

triangular number,

T_{n},

is the sum

1 + 2 + \dots + n .

We will find the explicit rule using the formula for finding the sum of an arithmetic series.

T_{n} = \frac{n ( a _{1} + a _{n} )}{2}

For this series, the first term is

a_{1} = 1

and the

n^{th}

term is

a_{n} = n .

Now, let's substitute the values and find the explicit rule for

T_{n} .

T_{n} = \frac{n ( a _{1} + a _{n} )}{2}

SubstituteII

$a_{1} = 1$ , $a_{n} = n$

T_{n} = \frac{n ( 1 + n )}{2}

Therefore, the explicit rule for the sequence of triangular numbers is

T_{n} = \frac{n ( n + 1 )}{2} .

Square Numbers

The $n^{th}$ square number, $S_{n},$ represents the number of dots in a square with the side made of $n$ dots. Therefore, the total number of dots is $n \cdot n = n^{2} .$ This tells us that the explicit rule for the sequence of square numbers is $S_{n} = n^{2} .$

b Of course, the first triangular number is

T_{1} = 1 .

Notice that the number of dots in

T_{n}

n

more than the number of dots in

T_{n - 1} .

Therefore, we can write a recursive rule for

T_{n} .

T_{1} = 1 T_{n} = T_{n - 1} + n for n > 1

Now, let's deal with the square numbers. The first square number is

S_{1} = 1 .

Notice that the number of dots in

S_{n}

(n - 1) + (n - 1) + 1 = 2 n - 1

more than the number of dots in

S_{n - 1} .

Therefore, we can write a recursive rule for

S_{n} .

S_{1} = 1 S_{n} = S_{n - 1} + 2 n - 1 for n > 1

c Of course

T_{1} = S_{1} .

Now, let's analyze the situation when

n > 1 .

Notice that a square

S_{n}

can be divide into two smaller triangles,

T_{n}

and

T_{n - 1},

as in the following picture.

This tells us that $S_{n} = T_{n} + T_{n - 1}$ for $n > 1 .$

Subchapter links

Communicate Your Answer p.441

Monitoring Progress p.442-446

Exercises p.447-450

Exercise 69 Page 450

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Triangular Numbers

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