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Use the Principle of Mathematical Induction to show that the following statement is true for all natural numbers [tex]\( n \)[/tex]:

[tex]\[ 6 + 12 + 18 + \cdots + 6n = 3n(n + 1) \][/tex]

What two conditions must the given statement satisfy to prove that it is true for all natural numbers? Select all that apply.

A. The statement is true for the natural number 1.
B. If the statement is true for some natural number [tex]\( k \)[/tex], it is also true for the next natural number [tex]\( k + 1 \)[/tex].
C. The statement is true for any two natural numbers [tex]\( k \)[/tex] and [tex]\( k + 1 \)[/tex].
D. If the statement is true for the natural number 1, it is also true for the next natural number 2.

Show that the first of these conditions is satisfied by evaluating the left and right sides of the given statement for the first natural number:

[tex]\[ 6 + 12 + 18 + \cdots + 6n = 3n(n + 1) \][/tex]

[tex]\[
\begin{aligned}
\text{Left side} & = 6 \\
\text{Right side} & = 3 \cdot 1 \cdot (1 + 1) = 6
\end{aligned}
\][/tex]

To show that the second condition is satisfied, write the given statement for [tex]\( k + 1 \)[/tex]:

[tex]\[ 6 + 12 + \cdots + 6k + 6(k + 1) = 3(k + 1)(k + 2) \][/tex]

(Simplify your answers. Type your answers in factored form.)

Sagot :

GinnyAnsweridn
To use the Principle of Mathematical Induction to prove the statement

[tex]\[ 6 + 12 + 18 + \cdots + 6n = 3n(n+1) \][/tex]

for all natural numbers [tex]\( n \)[/tex], we need to satisfy two conditions:

1. Base Case: Verify that the statement is true for the natural number 1.
2. Inductive Step: Show that if the statement is true for some natural number [tex]\( k \)[/tex], then it is also true for the next natural number [tex]\( k + 1 \)[/tex].

### Base Case
For [tex]\( n = 1 \)[/tex], consider the left side and the right side of the given statement:

[tex]\[ 6(1) = 3(1)(1 + 1) \][/tex]

Simplify both sides:

[tex]\[ 6 = 3 \cdot 1 \cdot 2 \][/tex]
[tex]\[ 6 = 6 \][/tex]

Since the left side equals the right side, the base case is satisfied.

### Inductive Step
Assume the statement is true for some natural number [tex]\( k \)[/tex]; that is,

[tex]\[ 6 + 12 + 18 + \cdots + 6k = 3k(k + 1) \][/tex]

We need to prove that the statement is true for [tex]\( k + 1 \)[/tex]. Start by writing the statement for [tex]\( k + 1 \)[/tex]:

[tex]\[ 6 + 12 + 18 + \cdots + 6k + 6(k + 1) = 3(k + 1)((k + 1) + 1) \][/tex]

The left-hand side is the sum up to [tex]\( 6k \)[/tex] plus the next term [tex]\( 6(k + 1) \)[/tex], so:

[tex]\[ 6 + 12 + 18 + \cdots + 6k + 6(k + 1) \][/tex]

Using the inductive hypothesis, we know that:

[tex]\[ 6 + 12 + 18 + \cdots + 6k = 3k(k + 1) \][/tex]

Therefore, the left-hand side is:

[tex]\[ 3k(k + 1) + 6(k + 1) \][/tex]

Factor out [tex]\( 3(k + 1) \)[/tex] from both terms:

[tex]\[ 3k(k + 1) + 6(k + 1) = 3(k + 1)(k + 2) \][/tex]

This matches the right-hand side of the statement for [tex]\( k + 1 \)[/tex]:

[tex]\[ 3(k + 1)(k + 2) \][/tex]

Hence, we have shown that if the statement is true for [tex]\( k \)[/tex], it is also true for [tex]\( k + 1 \)[/tex]. Therefore, by the principle of mathematical induction, the given statement is true for all natural numbers [tex]\( n \)[/tex].

So, the two conditions that must be satisfied to prove the statement for all natural numbers are:

1. The statement is true for the natural number 1.
2. If the statement is true for some natural number [tex]\( k \)[/tex], it is also true for the next natural number [tex]\( k + 1 \)[/tex].
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