Prove using mathematical induction: $1 \cdot 2 \cdot 3 + \ldots + n \cdot (n+1) \cdot (n+2) =  \frac{n(n+1)(n+2)(n+3)}{4}$ So I have the easy stuff done. However, I'm not sure how to go about doing the inductive step with such an awkward proof. :/
Statement: 
$$ 1 \cdot 2 \cdot 3 + … + n \cdot (n+1) \cdot (n+ 2) = \frac{n \cdot (n + 1) \cdot (n + 2) \cdot (n +3)}{4} .$$
Base step: $n = 1$: 
$$ 0  \cdot (0 + 1) \cdot (0 + 2) + 1 \cdot (1 + 1) \cdot (1 +2) = \frac{ 1 \cdot (1+ 1) \cdot (1 + 2) \cdot (1 + 3) }{4} .$$
$$ 0 \cdot (0 + 1) \cdot (0 + 2) +1 \cdot (1 + 1) \cdot (1 +2) = 0 + 6 = 6$$
$$ \frac{ 1 \cdot (1+ 1) \cdot (1 + 2) \cdot (1 + 3) }{4} = \frac{24}{ 4} = 6 .$$
$$6 = 6.$$
Inductive step:
$n = n + 1$
 A: For the record, the "correct" way to prove this identity is combinatorially. Written differently (after dividing both sides by $6$), it reads
$$ \sum_{k=3}^{n+2} \binom{k}{3} = \binom{n+3}{4}. $$
This is because each $4$-subset of $[n+3]$ is composed of its maximum $k+1$ and a $3$-subset of $[k]$.
A: Recommendation 1. Never, ever, write "$n=n+1$". It's just wrong.
Recommendation 2. It's easier if you state the inductive step in two parts: state the induction hypothesis explicitly, then state what you need to prove.
(Recommendation 2 is my strongest recommendation, especially if you are beginning induction and are having some difficulty; it helps organize your thoughts, it helps you keep straight what you are assuming vs. what you want to prove, it helps prevent confusion between the inductive step and the theorem as a whole. It's a chore, perhaps, but I really do strongly recommend doing it).
Recommendation 3. It is often useful to prevent confusion by stating the inductive step using a letter different from what you are trying to prove, so that your "induction hypothesis" doesn't get confused with the theorem to be proven; using a letter that is not being used for anything else can help. (It also prevents writing things that bring such horror to a trained mathematician like "$n=n+1$"...)  Here, the statement we are trying to prove uses $n$ for the variable and nothing else. To avoid getting confused, use $k$ for the induction hypothesis and $k+1$ for what you need to prove in the induction hypothesis.
This is purely aesthetical; it plays no mathematical role. It's just a way to prevent possible confusion, and nothing more.
Here, try this:
Inductive step.
Induction hypothesis: The result holds for $k$. That is, we assume that:

$$(1)(2)(3) + (2)(3)(4) + \cdots + (k)(k+1)(k+2) = \frac{k(k+1)(k+2)(k+3)}{4}$$
  is true.

To be proven: The result holds for $k+1$; that is, we need to prove that:

$$\small (1)(2)(3) + (2)(3)(4) + \cdots + (k+1)\bigl((k+1)+1\bigr)\bigl((k+1)+2\bigr) = \frac{(k+1)(k+2)(k+3)(k+4)}{4}.$$
  is true.

(The above is what I mean by "state the induction hypothesis explicitly, then state what you need to prove)
Okay let's try to prove that the second highlighted statement is true. With induction, we want to bring the Induction Hypothesis into play somehow; usually, by trying to think of the $k+1$ case as being "do the $k$-step and then do something else"; then we can use the induction hypothesis to simplify the "$k$-step".  This works very well here.
We have:
$$(1)(2)(3) + (2)(3)(4) + \cdots + (k+1)(k+2)(k+3).$$
In order to think about it as "the $k$-step and then a bit more", let's write the the $k$th summand as well as the $k+1$st:
$$(1)(2)(3) + (2)(3)(4) + \cdots + k(k+1)(k+2)+ (k+1)(k+2)(k+3).$$
Now, we can do the sum by first adding the first $k$ summands, and then adding $(k+1)(k+2)(k+3)$ to the result of that; so the entire sum we have is equal to:
$$
\Biggl( (1)(2)(3)+(2)(3)(4)+\cdots+k(k+1)(k+2)\Biggr) + (k+1)(k+2)(k+3).$$
Now, we may realize that the stuff inside the big parenthesis is precisely 
the sum about which the Induction Hypothesis tells us something. The Induction Hypothesis says that:
$$(1)(2)(3) + (2)(3)(4) + \cdots + (k)(k+1)(k+2) = \frac{k(k+1)(k+2)(k+3)}{4},$$
so we can replace that first parenthetical expression with $\frac{k(k+1)(k+2)(k+3)}{4}$. So, in summary, we have:
$$\begin{align*}
(1)(2)(3) &+ (2)(3)(4) + \cdots + k(k+1)(k+2)+ (k+1)(k+2)(k+3)\\
&=\Biggl( (1)(2)(3)+(2)(3)(4)+\cdots+k(k+1)(k+2)\Biggr) + (k+1)(k+2)(k+3)\\
&= \Biggl(\frac{k(k+1)(k+2)(k+3)}{4}\Biggr) + (k+1)(k+2)(k+3)
\end{align*}$$
with the first equality being just by associativity of the sum; and the second equality justified because we have applied the induction hypothesis. 

That is, we have simplified the sum of the first $k$ terms through the use of the induction hypothesis. 

Can you take it from here? 
A: $$1 \cdot 2 \cdot 3 + \cdots + n(n+1)(n+2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}.$$
Step 1: (Must prove that the given relation is true for $n=1$) 
$$p(1): 1\cdot 2\cdot 3=\frac{1\cdot (1+1)\cdot (1+2)\cdot(1+3)}{4} \Rightarrow 1\cdot 2\cdot 3= 1\cdot 2\cdot 3.$$
The base case is true.
Step 2: Suppose thaat the given relation is valid for $n=k$.
$$p(k): 1\cdot 2 \cdot 3 + \cdots + k(k+1)(k+2) = \frac{k(k + 1)(k + 2)(k +3)}{4}.$$
Step 3: Now we must prove that the given relation is correct for $n=k+1$.
$$p(k+1): 1\cdot 2 \cdot 3 + \cdots + (k+1)(k+2)(k+3) = \frac{(k+1)(k+2)(k + 3)(k +4)}{4}.$$
The above relation can be written:
$$\underbrace{1\cdot 2 \cdot 3 + \cdots + k(k+1)(k+2)} + (k+1)(k+2)(k+3) = \frac{(k+1)(k + 2)(k + 3)(k +4)}{4}.$$
The underlined partion in fact is the step 2. So we have
\begin{align*}
& 1\cdot 2 \cdot 3 + \cdots + k(k+1)(k+2) + (k+1)(k+2)(k+3) \\
 & = \frac{k(k + 1)(k + 2)(k +3)}{4} + (k+1)(k+2)(k+3) \\
 & = \frac{k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)}{4} \\
 & = \frac{(k+1)(k+2)(k+3)(k+4)}{4}.
\end{align*}
