How to prove that the statement $ 4+10+16 + \cdots + (6n-2) = n(3n+1)$ for all $n \ge 1$ using mathematical induction? I know you begin by establishing that it is true for $n=1$ which gives $6(1)-2 = 1(3\cdot1\cdot+1)$.
Then I replace each $n$ for a $k$, and I suppose that is true for $6k-2=k(3k+1)$.
But then the next step which is to prove that it is also true for $k+1$ and here is where I get confused. How do I proceed from here?
 A: Here's a slightly different approach and one I would recommend using in the future when proving summation identities. If you can represent a sum such as
$$
4+10+16+\cdots+(6n-2)\tag{1}
$$
using $\Sigma$-notation, then I would recommend doing so because it makes your induction argument look much nicer in general, and it also makes it easier to see where you are using the induction hypothesis. We can easily use $\Sigma$-notation to represent the sum in $(1)$, for the pattern is readily seen, and the general term (i.e., $6n-2$) is already given:
$$
\sum_{i=1}^n(6i-2)=\underbrace{4}_{i=1}+\underbrace{10}_{i=2}+\underbrace{16}_{i=3}+\cdots+\underbrace{(6n-2)}_{i=n}.
$$
Hence, the statement we are trying to prove for $n\geq 1$ becomes the following:
$$
\sum_{i=1}^n(6i-2)=n(3n+1).
$$
Proof. For $n\geq 1$, let $S(n)$ denote the statement
$$
S(n) : \sum_{i=1}^n(6i-2)=n(3n+1).
$$
Base case ($n=1$): $S(1)$ says that $\sum_{i=1}^1(6i-2)=6(1)-2=4=1(3(1)+1)$, and this is true. 
Inductive step $S(k)\to S(k+1)$: Fix some $k\geq 1$ and assume that 
$$
S(k) : \color{blue}{\sum_{i=1}^k(6i-2)=k(3k+1)}
$$
holds. To be shown is that $S(k+1)$ follows where
$$
S(k+1) : \color{green}{\sum_{i=1}^{k+1}(6i-2)=(k+1)(3k+4)}.
$$
Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\color{green}{\sum_{i=1}^{k+1}(6i-2)} &= \color{blue}{\sum_{i=1}^k(6i-2)}+[6(k+1)-2]\tag{by defn. of $\Sigma$}\\[1em]
&= \color{blue}{k(3k+1)}+[6(k+1)-2]\tag{by $S(k)$, ind. hyp.}\\[0.5em]
&= (3k^2+k)+(6k+4)\tag{expand}\\[0.5em]
&= 3k^2+7k+4\tag{simplify}\\[0.5em]
&= \color{green}{(k+1)(3k+4)},\tag{factor}
\end{align}
we end up at the right-hand side of $S(k+1)$, thereby completing the inductive step. 
By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$

It may not be your preference at first, but can you see how much cleaner and clearer the proof looks using $\Sigma$-notation?
A: $\forall n\geq 1, n \in \mathbb{Z}{+}$, let $P(n)$ denote the statement: $4+10+16+\dots+(6n-2)=n(3n+1).$
(Basis step)  
$$P(1): 6(1)-2=4\,\, \text{and} \,\,1(3(1)+1)=1(4)=4$$ is true.
(Inductive step) 
Suppose
$$P(k): 4+10+16+\dots+(6k-2)=k(3k+1)$$
holds for some integer $k\geq 1$.
(You need to show that $P(k+1)$ holds true, that is $(k+1)(3(k+1)+1)$)
$$\begin{align} k(3k+1) + (6(k+1)-2) &= 3k^2+k+6k+4 \\
&=3k^2 +7k +4 \\
&=3k^2 +3k +4k +4 \\
&=3k(k+1)+4(k+1)\\
&=(3k+4)(k+1)\\
&=(k+1)(3k+4)\\
&=(k+1)(3k+3+1)\\
&=(k+1)(3(k+1)+1)\\
\end{align}$$
thus, $P(k+1)$ holds, as to be shown.
