3.E: Number Patterns (Exercises)

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.

Answer

\(2n^2-n\).

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. \(1+ \displaystyle \frac{1}{2}+ \displaystyle \frac{1}{4}+\cdots + \displaystyle \frac{1}{2^{16}}+\displaystyle \frac{1}{2^{17}}\)

Answer

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,

\(S_ n = 1+3+5+7+9+···+197+199\)

\(S n = 199+197+195+193+191+···+3+1\)

By adding we get,

\((2S_n = 200+200+200+200+200+···+200+200\)

\(2S_n = 100(200)\)

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

\(S_n = (50)(200)\)

\(S_n = 10000\)

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 \( n \geq 1, 4+10+16+\dots+(6n-2)=n(3n+1)\)

Answer

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.

Answer

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

Exercise \(\PageIndex{8}\): Proof by induction

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

Answer

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.

Answer

\( \frac{n}{n+1}\).

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. \(a_n= \displaystyle {n \choose 3 } \)

Answer

1. \(n^2+2n+1\)
2. \(3 n^2 \)
3. \(n \choose 2\).