Answer
please see step-by-step
Work Step by Step
Principle of Mathematical Induction$:$
If both following 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 $n\in \mathbb{N}$.
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Let $P(n)$: $1+4+7+10+\displaystyle \cdots+(3n -2)=\frac{n(3n-1)}{2}$.
1.
$P(1)$: $1=\displaystyle \frac{1[3(1)-1]}{2}=\frac{1\cdot 2}{2}$
is true.
2.
Assume that $P(k)$ is true.
$1+4+7+\displaystyle \cdots+(3k-2)=\frac{k(3k-1)}{2}\quad(*)$
(is $P(k+1)$ true?)
Add the next, $(k+1)$th term of the LHS of (*) to both sides:
$LHS= 1+4+7+10+\cdots+(3k-2)+[3(k+1)-2]$
$RHS=\displaystyle \frac{k(3k-1)}{2}+[3(k+1)-2]$
$= \displaystyle \frac{k(3k-1)}{2}+3k+1$
$= \displaystyle \frac{k(3k-1)}{2}+\frac{6k+2}{2}$
$=\displaystyle \frac{3k^{2}-k+6k+2}{2}$
$=\displaystyle \frac{3k^{2}+5k+2}{2}$
... factor the numerator,
$... $factors of 3(2)=6 that add to 5 are 2 and 3...
... $3k^{2}+5k+2=3k^{2}+3k+2k+2=3k(k+1)+2(k+1)=(k+1)(3k+2)$
$RHS= \displaystyle \frac{(k+1)(3k+2)}{2}$
which can be written as
$=\displaystyle \frac{(k+1)[3(k+1)-1]}{2}$,
which results in $P(k+1)$ being true.
$(P(k)$ is true $\Rightarrow P(k+1)$ is true$)$
So by the Principle of Mathematical Induction,
$P(n)$ is true for all $n\in \mathbb{N}$.
