Answer
The steps are explained below.
Work Step by Step
The principle of mathematical induction states that if both of these conditions are satisfied,
1. $P(1)$ is true.
2. For any $k\in \mathbb{N}$, $P(k)$ is true $\Rightarrow$ $P(k+1)$ is true.
then
$P(n)$ is true for all $n\in \mathbb{N}$.
In the given question,
$P(n)$: $5+8+11+...+(3n +2)=\frac{n(3n+7)}{2}$.
First we check the truth of $P(1)$:
$P(1) = 1(3*1 + 7)/2 = 5 = R.H.S$
So, $P(1)$ is true.
Let us assume that $P(k)$ is true.
$5+8+11+...+(3k+2)=\frac{k(3k+7)}{2}\quad(a)$
Now we check if $P(k+1)$ is true.
So, we add the $(k+1)$th term to both sides of $a$
$LHS= 5+8+11+...+(3k+2)+[3(k+1)+2]$
$RHS=\displaystyle \frac{k(3k+7)}{2}+[3(k+1)+2]$
$=\displaystyle \frac{3k^{2}+7k+6k+10}{2}$
$=\displaystyle \frac{3k^{2}+13k+10}{2}$
$3k^{2}+13k+10=3k^{2}+3k+10k+10=3k(k+1)+10(k+1)=(k+1)(3k+10)$
$RHS= \displaystyle \frac{(k+1)(3k+10)}{2}$
which can be written as
$=\displaystyle \frac{(k+1)[3(k+1)+7]}{2}$,
so, $P(k+1)$ is true.
$(P(k)$ is true $\Rightarrow P(k+1)$ is true$)$
So by the Principle of Mathematical Induction,
$P(n)$ is true for all $n\in \mathbb{N}$.
