The first five triangular numbers are shown, where n represents the number of dots in the base of the figure,
and d(n) represents the total number of dots in the figure.
1 3
6
10
15
When n=1, there is 1 dot. When n = 2, there are 3 dots. When n=3, there are 6 dots. Notice that the total
number of dots d(n)Increases by n each time.
Use induction to prove that d (n) = n(n+1)
=
Part A
Prove the statement is true for n=1.
Type your answer in the box.

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MrRoyal MrRoyal · Answer 1 · 2022-02-17T19:56:41+01:00

A mathematical proof can be carried out by mathematical induction and by contradiction

The function is given as:

[tex]d(n) = \frac{n(n + 1)}{2}[/tex]

When n = 1, we have:

[tex]d(1) = \frac{1(1 + 1)}{2}[/tex]

[tex]d(1) = 2[/tex]

When n = k, the function becomes

[tex]d(k) = \frac{k(k + 1)}{2}[/tex]

When n = k + 1, the function becomes

[tex]d(k) + k + 1 = \frac{k(k + 1)}{2} + k + 1[/tex]

Open the bracket

[tex]d(k) + k + 1 = \frac{k^2 + k}{2} + k + 1[/tex]

Take the LCM

[tex]d(k) + k + 1 = \frac{k^2 + k + 2k + 2}{2}[/tex]

Factorize

[tex]d(k) + k + 1 = \frac{k(k + 1) + 2(k + 1)}{2}[/tex]

Factor out k + 1

[tex]d(k) + k + 1 = \frac{(k + 2)(k + 1)}{2}[/tex]

This gives

[tex]d(k + 1)= \frac{(k + 2)(k + 1)}{2}[/tex]

Because k + 1 satisfies the given function, then the function [tex]d(n) = \frac{n(n + 1)}{2}[/tex] has been proved by induction