Prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1)$ Using induction
prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1) \forall n \in \mathbb{N}$ 
Attempt:
Let $n =1$ so $3(1)-2 = 1$ and $\frac{1}{2}(3(1)-1)=1$
Assume true at $n=k$ so $3k-2 = \frac{k}{2}(3k-1)$
What do I do next? Here's where I'm stuck:
Let $n=k+1$ So $3(k+1) -2 = \frac{k+1}{2}(3(k+1)-1)$ 
 A: If Induction is not mandatory,
using the formula for summation of Arithmetic series, the sum is
$$\frac n2{(1+3n-2)}$$

For induction,
let $\sum_{r=1}^n(3r-2)=\dfrac n2(3n-1)$ holds true for $n=m$
$\implies\sum_{r=1}^m(3r-2)=\dfrac m2(3m-1)$
So,  $\sum_{r=1}^{m+1}(3r-2)=\sum_{r=1}^m(3r-1)+3(m+1)-1$
$=\dfrac m2(3m-1)+3(m+1)-2$
$=\dfrac{3m^2-m+6m+6-4}2=\dfrac{(m+1)[3(m+1)-1]}2$ 
Now, establish the base case i.e., for $n=1$
A: Notice that you have an aritmetic progression with initial term $1$ and difference $3$. So, $a_1 = 1$, and $a_k = 1 + 3(k-1)$. Then, $3n - 2 = 1 + 3(k-1) \implies k = n$, so $3n - 2$ is the $n$th term of the progression. This way: $$S_n = n\frac{a_1 + a_n}{2} = \frac{n}{2}(3n-1).$$
A: $$1+4+7+…+(3n-2) = \frac{n}{2}(3n-1)$$
Given identity is true For $n=1$
$$\frac{1\cdot(3-1)}{2}=1\tag{1}$$
Assume it's true for some integer $k$
$$1+4+7+…+(3k-2) = \frac{k}{2}(3k-1)\tag{2}$$
for $k+1$ .
 We have to show that
$$1+4+7+…+(3k-2)+(3(k+1)-2) = \frac{k+1}{2}(3(k+1)-1)$$
$$1+4+7+…+(3k-2)+(3k+1) = \frac{k+1}{2}(3k+2)\tag{3}$$
Adding $(3k+1)$ to LHS and RHS of $(2)$
 We get
$$\begin{align}
1+4+7+…+(3k-2)+(3k+1) & = \frac{k}{2}(3k-1)+(3k+1)\\
&= \frac{k(3k-1)+2(3k+1)}{2}\\
&= \frac{3k^2+5k+2}{2}\\
&= \frac{3k^2+3k+2k+2}{2}\\
1+4+7+…+(3k-2)+(3k+1) &= \frac{(3k+1)(k+2)}{2}\tag{4}\\
\end{align}$$
So, if our Identity is true for some integer $n=k$ Then it's also true for $n=k+1$, Since our identity is true for $n=1$ , Using principle of Mathematical Induction we can say that it's true for all $n\in\mathbb N$
Alternate
$$S=1+4+7+…+(3n-5)+(3n-2)\tag{1}$$
$$S=(3n-2)+(3n-5)+\cdots+7+4+1\tag{2}$$
By adding $1$ and $2$ We get,
$$2S=\underbrace{(3n-1)+(3n-1)+\cdots+(3n-1)+(3n-1)}_{\text{ n times }}=n(3n-1)$$
$$S=1+4+7+…+(3n-5)+(3n-2)=\frac{n}{2}(3n-1)$$
A: If you can avoid induction, you can use Gauss's trick:
$S=1+4+7+\cdots+(3n-8)+(3n-5)+(3n-2)$
$S=(3n-2)+(3n-5)+(3n-8)+\cdots+7+4+1$
$2S=(3n-1)+(3n-1)+\cdots+(3n-1)$
$2S=n(3n-1)$
$S = \frac{n}{2}(3n-1)$
