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.

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.

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

Answer

$1+3+5+7+9+···+197+199$

Notice that $1,3,5,7,\cdots$ are terms of a sequence. This is an Arithmetic Sequence because the difference between consecutive terms remains the terms throughout the 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 $.

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

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

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

Answer

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.

\[ a_n=pn+q \]

and obtain

\[ p+q=4,\qquad 2p+q=10. \]

Subtracting the first equation from the second gives $p=6$ and $q=-2$.

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

Check that the 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.

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

Exercise $\PageIndex{5}$: Tribonacci

Define

$ T_0=0,\quad T_1=0,\quad T_2=1, $

with

$ T_n=T_{n-1}+T_{n-2}+T_{n-3},\qquad n\geq3. $

Then explicitly ask either for $T_0,\ldots,T_{14}$ or for the first 15 terms beginning with $T_2$.

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 $n \geq 3, \, b_n=3b_{n-2}+2b_{n-3}.$

Find $b_3,b_4,b_5$ and $b_6$.
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 that the 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:

Assume 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.

Assume

$ S_k=\frac{k(3k-1)}2. $

Then

$ \begin{aligned} S_{k+1} &=S_k+(3k+1)\\ &=\frac{k(3k-1)}2+(3k+1)\\ &=\frac{3k^2-k+6k+2}{2}\\ &=\frac{3k^2+5k+2}{2}\\ &=\frac{(k+1)(3k+2)}2. \end{aligned} $

Since

$ 3(k+1)-1=3k+2, $

this establishes the required formula for $k+1$.

Therefore, by induction, the statement holds for all \ (n \ in \ mathbb {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

$ a_n=\frac{n+1}{n+2},\qquad n\geq1. $

Exercise $\PageIndex{10}$: Recognizing sequence

The sequence is

$ t_1=1,\quad t_2=3+5,\quad t_3=7+9+11,\ldots $

Assuming each term is the sum of the next block of consecutive odd numbers, with block lengths $1,2,3,\ldots$, Predict $n^{th}$ term. Explain your prediction.

Answer: $ t_n=n(n^2-n+1). $

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

Show that the perimeter of the design by joining $n$ hexagons, each with side length $s=2$ cm, in a row is $8n+4$ cm.

Exercise $\PageIndex{12}$: Pentagonal Numbers (cornered)

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

Exercise $\PageIndex{13}$: Square Pyramidal numbers

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

Exercise $\PageIndex{14}$: Difference

Compute the difference of each of the following sequences:

$a_n= n^3$
$a_n=n^{\underline{3}}$
$a_n= \displaystyle {n \choose 3 } $

Answer

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

Exercise $\PageIndex{15}$

1. Build a finite-difference table for $ 2, 9, 28, 65, 126,\ldots $ and determine a polynomial formula.

2. Explain why the first four terms of a sequence are not enough to prove the sequence rule.

Exercise $\PageIndex{16}$

For a sequence or discrete function $a_n = f(n)$, the forward difference operator $\Delta$ is defined by:
$
\Delta f(n) = f(n+1) - f(n)
$

Compute the finite difference $\Delta(n^2)$.

Answer

Using the definition of the difference operator:
$
\Delta(n^2) = (n+1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1
$
Notice that while $\frac{d}{dx}(x^2) = 2x$, the discrete difference includes an extra lower-order term $+1$. To obtain rules identical to polynomial differentiation in discrete calculus, we define the falling factorial:
$
n^{\underline{k}} = n(n-1)(n-2)\cdots(n-k+1)
$
which obeys the rule $\Delta (n^{\underline{k}}) = k \cdot n^{\underline{k-1}}$.

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

Exercise \(\PageIndex{2}\): Finite sum

Exercise \(\PageIndex{3}\): Proof by induction

Exercise \(\PageIndex{4}\): Proof by induction

Exercise \(\PageIndex{5}\): Tribonacci

Exercise \(\PageIndex{6}\): Proof by induction

Exercise \(\PageIndex{7}\): Quadratic Sequence

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

Exercise \(\PageIndex{9}\): Recognizing sequence

Exercise \(\PageIndex{10}\): Recognizing sequence

Exercise \(\PageIndex{11}\): Proof by induction

Exercise \(\PageIndex{12}\): Pentagonal Numbers (cornered)

Exercise \(\PageIndex{13}\): Square Pyramidal numbers

Exercise \(\PageIndex{14}\): Difference

Exercise \(\PageIndex{15}\)

Exercise \(\PageIndex{16}\)