Proving $\sum_{r=1}^n(6r-2)=n(3n+1)$ by induction

Question

A series is defined by $\sum\limits_{r=1}^n(6r-2)$.

Use the method of induction to prove that $S_n=n(3n+1)$.

I am at the induction step but I am struggling to rearrange $k(3k+1)+6(k+1)-2$ into the correct form.

Answer 1

$3k^2 + k + 6k + 6 - 2$

Hint:

$(3k^2 + 6k + 3) + (k + 1)$

Answer 2

Core part of induction step: \begin{align} \sum_{r=1}^{k+1}(6r-2) &= [6(k+1)-2]+\sum_{r=1}^k(6k-2)\tag{by defn. of $\Sigma$}\\[0.5em] &= 6k+6-2+k(3k+1)\tag{by ind. hyp.}\\[0.5em] &= 3k^2+7k+4\tag{simplify}\\[0.5em] &= (k+1)(3k+4)\tag{factor} \end{align}