Proving $6+12+18+24+...+6n=3n(n+1)$ by induction I'm learning prove by induction, but for some reason I just can't figure out the following example. I'll just work out the example and tell you guys where I'm stuck.
prove the following by induction $$6+12+18+24+...+6n=3n(n+1)$$

*

*prove basis step (n=1).
$$
6 = 6*1 = 3*1(1+1) = 6
$$


*induction step, assume Sk is true, n = k
$$
6 + 12 + 18 + 24 + ... + 6k = 3k(k+1)
$$


*induction step n = k + 1
$$
6 + 12 + 18 + 24 + ... + 6(k+1) = 3(k+1)(k+2)
$$


*bring Sk to both sides
$$
3k(k+1)+6(k+1) = 3(k+1)(k+2)
$$


*Simplify, I really have no idea how these equal eachother, what are the simplifying steps taken in this example? Any help would be greatly appreciated!
 A: I'll write out how I would prove this.
Prove $$6 + 12 + \cdots + 6n = 3n(n + 1).$$
Proof. We will first prove that $$1 + 2 + \cdots + n = \frac{n(n+1)}{2} \tag{1}$$ for all $n \in \mathbb{N}.$ Note that $$1 = \frac{1(1+1)}{2}.$$ Suppose $(1)$ holds when $n = k.$ If that is the case, then $$1 + 2 + \cdots + k = \frac{k(k + 1)}{2}. \tag{2}$$ We now add $k + 1$ to both sides to $(2)$ to get $$1 + 2 + \cdots + k + k + 1 = \frac{k(k + 1)}{2} + k + 1. \tag{3}$$ Doing some algebra on the right hand side of $(3),$ we see that $$\frac{k(k + 1)}{2} + k + 1 = \frac{k(k+1) + 2(k + 1)}{2} = \frac{(k + 1)(k + 2)}{2},$$ which implies that we have proved that $(1)$ holds for all $n \in \mathbb{N}.$ We now multiply both sides of $(1)$ by $6$ to get $$6 + 12 + \cdots + 6n = 3n(n + 1).$$
A: Steps 3 and 4 don't look right. You can't just assume:
$6 + 12 + 18 + 24 + ... + 6k + 6(k+1) = 3(k+1)(k+2)$ and manipulate it.
You need to prove that from the inductive assumption. So you need to use the following:
$6 + 12 + 18 + 24 + ... + 6k = 3k(k+1)$ to try and prove:
$6 + 12 + 18 + 24 + ... + 6k + 6(k+1) = 3(k+1)(k+2)$
So what you can do is manipulate the left hand side, and show that it comes out to the right hand side
$6 + 12 + 18 + 24 + ... + 6k + 6(k+1)$
$=3k(k+1) + 6(k+1)$ (this is substitution from our inductive assumption)
$=(3k+6)(k+1)$
$=3(k+2)(k+1)$
$=3(k+1)(k+2)$
So that shows the left hand side equals the right hand side. And that completes the proof.
A: Here is how I would do the inductive step in this specific case:
Inductive hypothesis: for some $k$, we have $\:6 + 12 + 18 + 24 + ... + 6k = 3k(k+1)$.
We have to deduce that $\:6 + 12 + 18 + 24 + ... + 6(k+1) = 3(k+1)(k+2)$.
Now, we can group the terms in the l.h.s. and use the inductive hypothesis:
$$(6 + 12 + 18 + 24 + ... + 6k)+6(k+1) = 3k(k+1)+6(k+1)=3(k+1)(k+2).$$
A: $6+12+18+24+\dots+6n=6(1+2+3+4+\dots+n)=6n(n+1)/2=3n(n+1)$.
The key is the well-known Gauß sum.  He did it as a school boy by pairing the first and last term, second and second to last, etc...
