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].