Chapter 11.7, Problem 3CYU

(a)

To determine

Find the next 5 triangular numbers.

Answer to Problem 3CYU

The next 5 triangular numbers: $3,6,10,15,21$

Explanation of Solution

Given:

A number is triangular if it can be represented visually by a triangular array.

The first triangular number is 1. Find the next 5 triangular numbers.

Concept Used:

The first triangle has just 1

Calculation:

The first triangle has just 1

The second triangle has another row with 2 extra dots, making 1+ 2 = 3

The third triangle has another row with 3 extra dots, making 1 + 2 + 3 =6 

The forth triangle has another row with 4 extra dots, making 1 + 2 + 3 +4 = 10

The fifth triangle has another row with 5 extra dots, making 1 + 2 + 3 + 4 +5 = 15

The fifth triangle has another row with 6 extra dots, making 1 + 2 + 3 + 4 +5 + 6 = 21

Thus, the next 5 triangular numbers: $3,6,10,15,21$

(b)

To determine

Write a formula for the nth Triangular number.

Answer to Problem 3CYU

The nth Triangular number is the nth term of the natural number: $a_n=\frac{n(n+1)}{2}$

Explanation of Solution

Given:

A number is triangular if it can be represented visually by a triangular array.

Write a formula for the nth Triangular number.

Concept Used:

The nth Triangular number is the nth term of the natural number: $a_n=\frac{n(n+1)}{2}$

Thus, the nth Triangular number is the nth term of the natural number: $a_n=\frac{n(n+1)}{2}$

(c)

To determine

Prove that the sum of the first n triangular numbers equals $\frac{n(n+1)(n+2)}{6}$

Explanation of Solution

Given:

A number is triangular if it can be represented visually by a triangular array.

Prove that the sum of the first n triangular numbers equals $\frac{n(n+1)(n+2)}{6}$

Concept Used:

The nth Triangular number is the nth term of the natural number: $\frac{n(n+1)}{2}=\frac{1}{2}·(n^2+n)$

Calculation:

The nth Triangular number is the nth term of the natural number: $\frac{n(n+1)}{2}=\frac{1}{2}·(n^2+n)$

Sum of the first n triangular numbers = $\frac{1}{2}·{\displaystyle \sum _{k=1}^n(k^2+k)}$

  $\begin{array}{l}\frac{1}{2}·{\displaystyle \sum _{k=1}^n(k^2+k)}\\ \frac{1}{2}[{\displaystyle \sum _{k=1}^n{k}^2}+{\displaystyle \sum _{k=1}^{n}k}\\ \frac{1}{2}[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}][{\displaystyle \sum _{k=1}^n{k}^2}=\frac{n(n+1)(2n+1)}{6}\&{\displaystyle \sum _{k=1}^{n}k}=\frac{n(n+1)}{2}]\\ \frac{1}{2}·\frac{n(n+1)}{2}[\frac{(2n+1)}{3}+1][\text{Factor with }\frac{n(n+1)}{2}]\\ \frac{n(n+1)}{4}[\frac{(2n+1)}{3}+\frac{3}{3}]=\frac{n(n+1)}{4}·\frac{2n+4}{3}=\frac{n(n+1)}{4}·\frac{2(n+2)}{3}=\frac{n(n+1)(n+2)}{6}\\ \frac{1}{2}·{\displaystyle \sum _{k=1}^n(k^2+k)}=\frac{n(n+1)(n+2)}{6}\text{Proved}\end{array}$

Thus, the sum of the first n triangular numbers equals $\frac{n(n+1)(n+2)}{6}$ proved.