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 $3(k+1) -2 = \frac{k+1}{2}(3(k+1)-1)$ 

If you can avoid induction, you can use Gauss's trick:
$S=1+4+7+\cdots+(3n-8)+(3n-5)+(3n-2)$
$S=(3n-2)+(3n-5)+(3n-8)+\cdots+7+4+1$
$2S=(3n-1)+(3n-1)+\cdots+(3n-1)$
$2S=n(3n-1)$
$S = \frac{n}{2}(3n-1)$
