Prove the following by using the principle of mathematical induction for all $n \in N:$
$3 \times 6 + 6 \times 9 + 9 \times 12 + \ldots + (3n)(3n + 3) = 3n(n + 1)(n + 2)$
$3 \times 6 + 6 \times 9 + 9 \times 12 + \ldots + (3n)(3n + 3) = 3n(n + 1)(n + 2)$
(N/A) Let $P(n)$ be the statement: $3 \times 6 + 6 \times 9 + 9 \times 12 + \ldots + (3n)(3n + 3) = 3n(n + 1)(n + 2)$.
Step $1$: For $n = 1$,the $LHS$ is $3 \times 6 = 18$. The $RHS$ is $3(1)(1 + 1)(1 + 2) = 3 \times 2 \times 3 = 18$. Since $LHS$ = $RHS$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$3 \times 6 + 6 \times 9 + \ldots + (3k)(3k + 3) = 3k(k + 1)(k + 2)$.
Step $3$: We need to show $P(k + 1)$ is true,i.e.,$3 \times 6 + \ldots + (3k)(3k + 3) + (3(k + 1))(3(k + 1) + 3) = 3(k + 1)(k + 2)(k + 3)$.
Adding $(3(k + 1))(3(k + 1) + 3)$ to both sides of the assumption:
$LHS$ = $3k(k + 1)(k + 2) + (3k + 3)(3k + 6)$
= $3k(k + 1)(k + 2) + 3(k + 1) \times 3(k + 2)$
= $3(k + 1)(k + 2) [k + 3]$
= $3(k + 1)(k + 2)(k + 3)$.
Thus,$P(k + 1)$ is true. By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.
Step $1$: For $n = 1$,the $LHS$ is $3 \times 6 = 18$. The $RHS$ is $3(1)(1 + 1)(1 + 2) = 3 \times 2 \times 3 = 18$. Since $LHS$ = $RHS$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$3 \times 6 + 6 \times 9 + \ldots + (3k)(3k + 3) = 3k(k + 1)(k + 2)$.
Step $3$: We need to show $P(k + 1)$ is true,i.e.,$3 \times 6 + \ldots + (3k)(3k + 3) + (3(k + 1))(3(k + 1) + 3) = 3(k + 1)(k + 2)(k + 3)$.
Adding $(3(k + 1))(3(k + 1) + 3)$ to both sides of the assumption:
$LHS$ = $3k(k + 1)(k + 2) + (3k + 3)(3k + 6)$
= $3k(k + 1)(k + 2) + 3(k + 1) \times 3(k + 2)$
= $3(k + 1)(k + 2) [k + 3]$
= $3(k + 1)(k + 2)(k + 3)$.
Thus,$P(k + 1)$ is true. By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.
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