# 3.E: Number Patterns (Exercises)

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## Exercise $$\PageIndex{1}$$: Hexagonal numbers (cornered)

Consider the hexagonal numbers are the sequence $$1,6,15, 28,45,66 \cdots.$$ Predict the n th term. Explain your prediction.

$$2n^2-n$$.

## Exercise $$\PageIndex{2}$$: Finite sum

For each of the following, find the sum and explain your reasoning. Please do not use any formula.

1. $$1+3+5+7+9+\cdots +197+199$$
2. $$1+ \displaystyle \frac{1}{2}+ \displaystyle \frac{1}{4}+\cdots + \displaystyle \frac{1}{2^{16}}+\displaystyle \frac{1}{2^{17}}$$
1. $$1+3+5+7+9+···+197+199$$

Notice that $$1,3,5,7 ,\cdots$$ terms of a sequence. This is an Arithmetic Sequence because the difference remains the same between the terms throughout the entire sequence. Hence,$$a = 1 \& \, d = 2$$.

Consider,

$$S_ n = 1+3+5+7+9+···+197+199$$

$$S n = 199+197+195+193+191+···+3+1$$

$$(2S_n = 200+200+200+200+200+···+200+200$$

$$2S_n = 100(200)$$

$$S_n = ((100)/2))(200)$$

$$S_n = (50)(200)$$

$$S_n = 10000$$

Hence, the sum of the sequence is $$10000.$$

2.

## Exercise $$\PageIndex{3}$$: Proof by induction

Consider the sequence $$4,10,16, 22, 28,,\dots$$, assume that the pattern continues.

1. Show that the $$n^{th}$$ term of this sequence can be expressed as $$6n-2$$.
2. Prove by using induction for all integers $$n \geq 1, 4+10+16+\dots+(6n-2)=n(3n+1)$$

1.

 Term First difference 4 10 6 16 6 22 6

Notice that the first difference is constant. Hence the $$n^{th}$$ term is a linear function.

Let $$t_n = a_n + b.$$

Then we need to find $$a, b$$.

First Equation: Let $$n = 1$$

$$t_1 = 4$$

$$4 = a(1) + b$$

$$4 = a + b$$

Second Equation: Let $$n = 2$$

$$t_2 = 10$$

$$10 = a(2) + b$$

$$10 = 2a + b$$

To find $$a$$, we use $$10 = 2a + b$$ and $$- 4 = a + b$$. Therfore, $$6 = a.$$

Now to find $$b$$, we use $$a = 6$$ and $$4 = a + b$$,

$$4 = (6) + b$$

$$4 - 6 = b$$

$$-2 = b$$.

Hence,$$t_n = 6n - 2.$$

2. Step 1: Base Step: Show that this statement is true for the smallest value

Check statement is true for n = 1.

L.H.S = 4

R.H.S = n(3n + 1)

= (1)(3(1) + 1)

= (1)(3 + 1)

= (1)(4)

= 4

Hence, the statement is true for n = 1.

Step 2: Induction Assumption:

We shall assume that the statement is true for n = k.

4 + 10 + 16 + . . . + (6k − 2) = k(3k + 1)

Step 3: Induction:

We shall show that the statement is true for n = k + 1.

4 + 10 + 16 + . . . + (6k − 2) + (6 (k + 1) − 2) = (k+1)(3(k + 1) + 1)

Consider, L.H.S = 4 + 10 + 16 + . . . + (6k − 2) + (6 (k + 1) − 2)

= k(3k + 1) + (6 ( k + 1) - 2)

= k (3k + 1) + (6k + 6 - 2)

= k (3k + 1) + (6k + 4)

= 3k 2 + k + 6k + 4

= 3k 2 + 7k + 4

= (k + 1)(3k + 4)

Hence, the statement is true for n = k + 1

Therefore, by induction the statement is true, ∀n ∈ N.

## Exercise $$\PageIndex{4}$$: Proof by induction

Consider the sequence $$3,11,19, 27, 35,\dots$$, assume that the pattern continues.

1. Show that the $$n^{th}$$ term of this sequence can be expressed as $$8n-5$$.
2. Prove by using induction for all integers $$n\geq 1, 3+11+19 \dots + (8n-5)=4n^2-n.$$

## Exercise $$\PageIndex{5}$$: Tribonacci

Let's start with the numbers $$0,0,1,$$ and generate future numbers in our sequence by adding up the previous three numbers. Write out the first $$15$$ terms in this sequence, starting with the first $$1$$.

## Exercise $$\PageIndex{6}$$: Proof by induction

The sequence $$b_0,b_1,b_2....$$ is defined as follows: $$b_0=1,b_1=3,b_2=5,$$ and for any integer $$n \geq 3, \, b_n=3b_{n-2}+2b_{n-3}.$$

1. Find $$b_3,b_4,b_5$$ and $$b_6$$.
2. Prove that $$b_n < 2^{n+1}$$ for all integers $$n \geq 1.$$

## Exercise $$\PageIndex{7}$$: Quadratic Sequence

Find the $$n^{th}$$ term of the sequence $$5,10,17, 26, 37, \cdots$$, assume that the pattern continues.

$$(n+1)^2+1=n^2+2n+2$$

## Exercise $$\PageIndex{8}$$: Proof by induction

Prove by using induction: for all integers $$n\geq 1, \, 1+4+7 \dots + (3n-2)=\frac{n(3n-1)}{2}.$$

Step 1: Base Step: Show that this statement is true for the smallest value

Check statement is true for n = 1.

L.H.S = 1

R.H.S = n(3n−1) / (2)

= (1)(3(1) − 1) / (2)

= (1)(3 − 1) / (2)

= (1)(2) / (2)

= 1

Hence, the statement is true for n = 1.

Step 2: Induction Assumption:

We shall assume that the statement is true for n = k.

1+4+7...+(3k−2)= k(3k−1) / (2)

Step 3: Induction:

We shall show that the statement is true for n = k + 1.

1+4+7...+ (3k - 2) + (3(k + 1) − 2) = (k + 1)(3(k + 1) − 1) / (2)

Consider, L.H.S = k(3k − 1) / (2) + (3 (k + 1) − 2)

= k (3k − 1) / (2) + (3k + 3) − 2)

= k (3k − 1) / (2) + (3k + 1)

= (3k 2 + k ) / (2) + (3k + 1)

= (3k 2 + k + 3k + 1) / (2)

= (3k 2 + 4k + 1) / (2)

= ((k + 1)(3k + 1)) / (2)

Hence, the statement is true for n = k + 1

Therefore, by induction the statement is true, ∀n ∈ N.

## Exercise $$\PageIndex{9}$$: Recognizing sequence

Predict $$n^{th}$$ term of the sequence $$\frac{2}{3},\frac{3}{4}, \frac{4}{5}\cdots\,$$ assume that the pattern continues. Explain your prediction.

$$\frac{n}{n+1}$$.

## Exercise $$\PageIndex{10}$$: Recognizing sequence

Consider the sequence $$t_1=1, t_2=3+5, t_3=7+9+11, \cdots$$. Predict the n th term. Justify your prediction.

## Exercise $$\PageIndex{11}$$: Proof by induction

Show that the perimeter of the design by joining $$n$$ hexagons in a row is $$8n+4$$ cm.

## Exercise $$\PageIndex{13}$$: Pentagonal Numbers (cornered)

Find the $$n^{th}$$ term of the sequence $$1,5,12, 22, \cdots$$,assume that the pattern continues.

## Exercise $$\PageIndex{14}$$: Square Pyramidal numbers

Find the $$n^{th}$$ term of the sequence $$1,5,14,30 \cdots$$., assume that the pattern continues.

## Exercise $$\PageIndex{15}$$: Difference

Compute the difference of each of the following sequences:

1. $$a_n= n^3$$
2. $$a_n=n^{\underline{3}}$$
3. $$a_n= \displaystyle {n \choose 3 }$$
1. $$n^2+2n+1$$
2. $$3 n^2$$
3. $$n \choose 2$$.