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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]

1. Base Case:

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]\(n = 1\)[/tex].

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

2. Induction Hypothesis:

Assume that the statement is true for some natural number [tex]\(k\)[/tex], i.e.,

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

3. Inductive Step:

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

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

Simplify the left side using the induction hypothesis:

[tex]\[
\begin{aligned}
3k(k+1) + 6(k+1) & = 3(k+1)(k+2)
\end{aligned}
\][/tex]

If the statement for [tex]\(k + 1\)[/tex] is true whenever the given statement [tex]\(6 + 12 + 18 + \cdots + 6n = 3n(n+1)\)[/tex] is true for all natural numbers.

4. Conclusion:

Using the statement for [tex]\(k\)[/tex]:

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

To simplify the left side for [tex]\(k + 1\)[/tex]:

[tex]\[
\begin{aligned}
3k(k+1) + 6(k+1) &= 3(k+1)(k+2) \\
3(k+1)(k+1 + 1) &= 3(k+1)(k+2)
\end{aligned}
\][/tex]

Thus, by the principle of mathematical induction, the statement is true for all natural numbers [tex]\(n\)[/tex].



Answer :

GinnyAnswer
To use the Principle of Mathematical Induction to prove that the statement [tex]\( 6 + 12 + 18 + \cdots + 6n = 3n(n+1) \)[/tex] is true for all natural numbers [tex]\( n \)[/tex], we need to follow two steps:

1. Base Case: Verify the statement for [tex]\( n = 1 \)[/tex].
2. Induction Step: Assume the statement holds for some [tex]\( k \)[/tex] (induction hypothesis), and then prove it for [tex]\( k + 1 \)[/tex].

### Step 1: Base Case

Evaluate the left and right sides of the given statement for the first natural number [tex]\( n = 1 \)[/tex].

#### Left Side:
[tex]\[ 6 \cdot 1 = 6 \][/tex]

#### Right Side:
[tex]\[ 3 \cdot 1 \cdot (1 + 1) = 3 \cdot 1 \cdot 2 = 6 \][/tex]

Both sides are equal, so the base case holds:
[tex]\[ 6 = 6 \][/tex]

### Step 2: Induction Step

Assume the statement is true for [tex]\( n = k \)[/tex]. That is, we assume:
[tex]\[ 6 + 12 + 18 + \cdots + 6k = 3k(k+1) \][/tex]

We need to show that under this assumption, the statement is also true for [tex]\( n = k + 1 \)[/tex].

Write the given statement for [tex]\( n = k + 1 \)[/tex]:
[tex]\[ 6 + 12 + 18 + \cdots + 6k + 6(k+1) = 3(k+1)((k+1)+1) \][/tex]

Simplify the right side:
[tex]\[ 3(k+1)(k+2) \][/tex]

For the left side, use the induction hypothesis:
[tex]\( 6 + 12 + 18 + \cdots + 6k = 3k(k+1) \)[/tex].

So the left side for [tex]\( k + 1 \)[/tex] becomes:
[tex]\[ 3k(k+1) + 6(k+1) \][/tex]

Factor out [tex]\( 3(k+1) \)[/tex] from the terms on the left side:
[tex]\[ 3(k+1)k + 6(k+1) = 3(k+1)(k + 2) \][/tex]

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

We see that the left side [tex]\( 3(k+1)(k + 2) \)[/tex] matches the simplified right side for [tex]\( k + 1 \)[/tex]:
[tex]\[ 3(k+1)(k+2) \][/tex]

Since both sides are equal, the induction step holds. Therefore, if the statement is true for [tex]\( n = k \)[/tex], it is also true for [tex]\( n = k + 1 \)[/tex].

### Conclusion

By the Principle of Mathematical Induction, since the statement is true for the base case ([tex]\( n = 1 \)[/tex]) and the induction step holds, the statement [tex]\( 6 + 12 + 18 + \cdots + 6n = 3n(n+1) \)[/tex] is true for all natural numbers [tex]\( n \)[/tex].

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