<s>[INST] 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:
$n = n + 1$
 [/INST] $$1 \cdot 2 \cdot 3 + \cdots + n(n+1)(n+2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}.$$
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*}
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